街头斗殴数学

Street-Fighting Mathematics

© 2010 Sanjoy Mahajan

© 2010 by Sanjoy Mahajan

前言 © 2010 Carver A. Mead

Foreword © 2010 by Carver A. Mead

《街头斗殴数学：有根据的猜测和机会主义问题解决的艺术》由 Sanjoy Mahajan（作者）、Carver A. Mead（序言）和麻省理工学院出版社（出版商）编写，根据知识共享署名-非商业性-相同方式共享 3.0 美国许可证授权。

Street-Fighting Mathematics: The Art of Educated Guessing and Opportunistic Problem Solving by Sanjoy Mahajan (author), Carver A. Mead (foreword), and MIT Press (publisher) is licensed under the Creative Commons Attribution–Noncommercial–Share Alike 3.0 United States License.

许可证副本可在http://creativecommons.org/licenses/by-nc-sa/3.0/us/获取。

A copy of the license is available at http://creativecommons.org/licenses/by-nc-sa/3.0/us/

有关特殊数量折扣的信息，请发送电子邮件至 special_sales@mitpress.mit.edu

For information about special quantity discounts, please email special_sales@mitpress.mit.edu

作者使用 ConT _E Xt 和 PDFT _E X在 Palatino 和 Euler 中排版

Typeset in Palatino and Euler by the author using ConT_EXt and PDFT_EX

国会图书馆在版数据​ ​​​​​​​

LIBRARY OF CONGRESS CATALOGING-IN-PUBLICATION DATA

马哈詹，桑乔伊，1969–

Mahajan, Sanjoy, 1969–

街头斗殴数学：有根据的猜测和机会主义问题解决的艺术 / Sanjoy Mahajan；Carver A. Mead 作序。

Street-fighting mathematics : the art of educated guessing and opportunistic problem solving / Sanjoy Mahajan ; foreword by Carver A. Mead.

     页厘米

     p. cm.

包括参考书目和索引。

Includes bibliographical references and index.

ISBN 978-0-262-51429-3（平装：碱性纸）

ISBN 978-0-262-51429-3 (pbk. : alk. paper)

ISBN 978-0-262-26559-1（电子书）

ISBN 978-0-262-26559-1 (e-book)

1. 问题解决。2. 假设。3. 估计理论。I. 标题。

1. Problem solving. 2. Hypothesis. 3. Estimation theory. I. Title.

QA63.M34 2010

QA63.M34 2010

510—dc22

510—dc22

2009028867

2009028867

美国印刷装订

Printed and bound in the United States of America

10 9 8 7 6 5 4 3 2 1

10  9  8  7  6  5  4  3  2  1

致朱丽叶

For Juliet

简要内容

Brief contents

前言

Foreword

前言

Preface

1 尺寸

1    Dimensions

2个简单案例

2    Easy cases

3 合并

3    Lumping

4 图片证明

4    Pictorial proofs

5 取出大部件

5    Taking out the big part

6 类比

6    Analogy

参考书目

Bibliography

指数

Index

内容

Contents

前言

Foreword

前言

Preface

1 尺寸

1    Dimensions

1.1 经济学：跨国公司的力量

1.1 Economics: The power of multinational corporations

1.2 牛顿力学：自由落体

1.2 Newtonian mechanics: Free fall

1.3 猜测积分

1.3 Guessing integrals

1.4总结和进一步的问题

1.4 Summary and further problems

2个简单案例

2    Easy cases

2.1 重温高斯积分

2.1 Gaussian integral revisited

2.2 平面几何：椭圆的面积

2.2 Plane geometry: The area of an ellipse

2.3 立体几何：截头金字塔的体积

2.3 Solid geometry: The volume of a truncated pyramid

2.4 流体力学：阻力

2.4 Fluid mechanics: Drag

2.5总结和进一步的问题

2.5 Summary and further problems

3 合并

3    Lumping

3.1 估计人口：有多少婴儿？

3.1 Estimating populations: How many babies?

3.2 估算积分

3.2 Estimating integrals

3.3 估计导数

3.3 Estimating derivatives

3.4 分析微分方程：弹簧质量系统

3.4 Analyzing differential equations: The spring–mass system

3.5 预测单摆的周期

3.5 Predicting the period of a pendulum

3.6总结和进一步的问题

3.6 Summary and further problems

4 图片证明

4    Pictorial proofs

4.1 奇数加法

4.1 Adding odd numbers

4.2 算术平均数和几何平均数

4.2 Arithmetic and geometric means

4.3 近似对数

4.3 Approximating the logarithm

4.4 平分三角形

4.4 Bisecting a triangle

4.5 求和级数

4.5 Summing series

4.6总结和进一步的问题

4.6 Summary and further problems

5 取出大部件

5    Taking out the big part

5.1 使用“一”和“少”进行乘法

5.1 Multiplication using one and few

5.2 分数变化和低熵表达式

5.2 Fractional changes and low-entropy expressions

5.3 具有一般指数的分数变化

5.3 Fractional changes with general exponents

5.4 逐次逼近：井有多深？

5.4 Successive approximation: How deep is the well?

5.5 令人畏惧的三角积分

5.5 Daunting trigonometric integral

5.6总结和进一步的问题

5.6 Summary and further problems

6 类比

6    Analogy

6.1 空间三角学：甲烷中的键角

6.1 Spatial trigonometry: The bond angle in methane

6.2 拓扑：有多少个区域？

6.2 Topology: How many regions?

6.3 算子：欧拉-麦克劳林求和

6.3 Operators: Euler–MacLaurin summation

6.4 正切根：一个令人畏惧的超越和

6.4 Tangent roots: A daunting transcendental sum

6.5一路顺风

6.5 Bon voyage

参考书目

Bibliography

指数

Index

前言

Foreword

我们大多数人都从数学家那里学习数学课程——坏主意！

Most of us took mathematics courses from mathematicians—Bad Idea!

数学家视数学为一门独立的学科。我们其他人则将数学作为一种精确的语言，用来表达现实世界中量之间的关系，并以此作为从这些关系中得出定量结论的工具。然而，如今教授的数学课程却很少能起到这样的作用，甚至常常带来负面影响。

Mathematicians see mathematics as an area of study in its own right. The rest of us use mathematics as a precise language for expressing relationships among quantities in the real world, and as a tool for deriving quantitative conclusions from these relationships. For that purpose, mathematics courses, as they are taught today, are seldom helpful and are often downright destructive.

作为一名学生，我曾承诺过自己，如果我将来成为一名老师，我绝不会让学生接受那种教学方式。我一生都在努力寻找直接透明地观察现实的方法，并试图将这些洞见量化地表达出来，我从未故意违背过我的诺言。

As a student, I promised myself that if I ever became a teacher, I would never put a student through that kind of teaching. I have spent my life trying to find direct and transparent ways of seeing reality and trying to express these insights quantitatively, and I have never knowingly broken my promise.

除了极少数例外，我发现最有用的数学知识都是在科学和工程课上、自学或从本书中学到的。《街头斗殴数学》是一股清风。Sanjoy Mahajan 以最友好的方式向我们传授了在现实世界中行之有效的工具。就在我们认为某个主题显而易见的时候，他又将我们带入了另一个层次。我个人最喜欢的是纳维-斯托克斯方程的解法：它太难了，我甚至从未尝试过求解。但他带领我们解决了这个问题，并一路收获了宝贵的洞见。

With rare exceptions, the mathematics that I have found most useful was learned in science and engineering classes, on my own, or from this book. Street-Fighting Mathematics is a breath of fresh air. Sanjoy Mahajan teaches us, in the most friendly way, tools that work in the real world. Just when we think that a topic is obvious, he brings us up to another level. My personal favorite is the approach to the Navier–Stokes equations: so nasty that I would never even attempt a solution. But he leads us through one, gleaning gems of insight along the way.

这本小书里蕴含着我们每个人的真知灼见。本书中的一些技巧，我亲自实践过，也推荐给大家。

In this little book are insights for every one of us. I have personally adopted several of the techniques that you will find here. I recommend it highly to every one of you.

—卡弗·米德

—Carver Mead

前言

Preface

过度的数学严谨性会导致僵硬：即使得出了正确的结果，也害怕做出不合理的猜测。与其麻痹大意，不如鼓起勇气——先行动，后提问。虽然作为公共政策并不明智，但这却是一种宝贵的解决问题的哲学，也是本书的主题：如何在没有证明或精确计算的情况下猜测答案。

Too much mathematical rigor teaches rigor mortis: the fear of making an unjustified leap even when it lands on a correct result. Instead of paralysis, have courage—shoot first and ask questions later. Although unwise as public policy, it is a valuable problem-solving philosophy, and it is the theme of this book: how to guess answers without a proof or an exact calculation.

有根据的猜测和随机应变的问题解决需要一个工具箱。用乔治·波利亚的话来说，工具是我使用过两次的技巧。本书构建、完善并演示了跨人类知识不同领域的实用工具。丰富的示例有助于将工具（一般原理）与具体应用区分开来，以便你能够掌握工具并将其迁移到你感兴趣的问题中。

Educated guessing and opportunistic problem solving require a toolbox. A tool, to paraphrase George Polya, is a trick I use twice. This book builds, sharpens, and demonstrates tools useful across diverse fields of human knowledge. The diverse examples help separate the tool—the general principle—from the particular applications so that you can grasp and transfer the tool to problems of particular interest to you.

用于教授该工具的例子包括：不积分地猜测积分、反驳媒体中的常见论点、从非线性微分方程中提取物理性质、不解纳维-斯托克斯方程地估计阻力、寻找平分三角形的最短路径、猜测键角、以及对每个项都是未知和超越的无穷级数求和。

The examples used to teach the tools include guessing integrals without integrating, refuting a common argument in the media, extracting physical properties from nonlinear differential equations, estimating drag forces without solving the Navier–Stokes equations, finding the shortest path that bisects a triangle, guessing bond angles, and summing infinite series whose every term is unknown and transcendental.

本书是对《如何解决问题》 [ 37 ]、《数学与合理推理》 [ 35,36 ]和《问题解决的艺术与技巧》 [ 49 ]等著作的补充。这些著作教导我们如何精确地解决精确陈述的问题，而生活中我们经常遇到一些定义不明确的问题，只需要中等精度的解决方案。精度仅为2倍的计算可能表明，拟建的桥梁永远不会建成，或者电路永远无法正常工作。省去进行精确分析所节省的精力，可以用来发明有前景的新设计。

This book complements works such as How to Solve It [37], Mathematics and Plausible Reasoning [35, 36], and The Art and Craft of Problem Solving [49]. They teach how to solve exactly stated problems exactly, whereas life often hands us partly defined problems needing only moderately accurate solutions. A calculation accurate only to a factor of 2 may show that a proposed bridge would never be built or a circuit could never work. The effort saved by not doing the precise analysis can be spent inventing promising new designs.

本书源于我在麻省理工学院教授的同名短期课程，该课程持续了数年。学生的经历各不相同：从本科一年级到准备从事研究和教学工作的研究生。学生的专业领域也各不相同：从物理、数学、管理学到电气工程、计算机科学和生物学，应有尽有。尽管如此，或者正因为如此，学生们似乎受益于这套工具，并享受着插图和应用的多样性。祝愿你们也一样。

This book grew out of a short course of the same name that I taught for several years at MIT. The students varied widely in experience: from first-year undergraduates to graduate students ready for careers in research and teaching. The students also varied widely in specialization: from physics, mathematics, and management to electrical engineering, computer science, and biology. Despite or because of the diversity, the students seemed to benefit from the set of tools and to enjoy the diversity of illustrations and applications. I wish the same for you.

如何使用本书

How to use this book

亚里士多德曾是年轻的马其顿国王亚历山大（后来的亚历山大大帝）的导师。正如古代皇室所知，一位技艺精湛、知识渊博的导师是最有效的教师[ 8 ]。一位技艺精湛的导师少言寡语，多提问，因为她深知提问、思考和讨论能够促进持久的学习。因此，本书穿插了两种类型的问题。

Aristotle was tutor to the young Alexander of Macedon (later, Alexander the Great). As ancient royalty knew, a skilled and knowledgeable tutor is the most effective teacher [8]. A skilled tutor makes few statements and asks many questions, for she knows that questioning, wondering, and discussing promote long-lasting learning. Therefore, questions of two types are interspersed through the book.

页边空白处标有 的问题：这些问题是导师在辅导课上可能会问你的，并要求你制定分析的后续步骤。这些问题的答案在后续文本中，你可以在其中查看你的答案和我的分析。

Questions marked with a in the margin: These questions are what a tutor might ask you during a tutorial, and ask you to work out the next steps in an analysis. They are answered in the subsequent text, where you can check your solutions and my analysis.

编号问题：这些问题用阴影背景标记，是导师在辅导课后可能会给你带回家的。它们要求你练习工具，扩展示例，同时使用多个工具，甚至解决（明显的）悖论。

Numbered problems: These problems, marked with a shaded background, are what a tutor might give you to take home after a tutorial. They ask you to practice the tool, to extend an example, to use several tools together, and even to resolve (apparent) paradoxes.

尝试两种类型的许多问题！

Try many questions of both types!

版权许可

Copyright license

本书采用与麻省理工学院开放课程相同的许可协议：知识共享署名-非商业性使用-相同方式共享许可协议。出版商和我鼓励您以非商业形式使用、改进和分享本书，我们乐意接受任何更正和建议。

This book is licensed under the same license as MIT’s OpenCourseWare: a Creative Commons Attribution-Noncommercial-Share Alike license. The publisher and I encourage you to use, improve, and share the work non-commercially, and we will gladly receive any corrections and suggestions.

致谢

Acknowledgments

我衷心感谢以下个人和组织。

I gratefully thank the following individuals and organizations.

标题：卡尔·莫耶 (Carl Moyer)。

For the title: Carl Moyer.

编辑指导： Katherine Almeida 和 Robert Prior。

For editorial guidance: Katherine Almeida and Robert Prior.

对手稿进行全面、透彻的评论： Michael Gottlieb、David Hogg、David MacKay 和 Carver Mead。

For sweeping, thorough reviews of the manuscript: Michael Gottlieb, David Hogg, David MacKay, and Carver Mead.

为激励大家而设的老师有： John Allman、Arthur Eisenkraft、Peter Goldreich、John Hopfield、Jon Kettenring、Geoffrey Lloyd、Donald Knuth、Carver Mead、David Middlebrook、Sterl Phinney 和 Edwin Taylor。

For being inspiring teachers: John Allman, Arthur Eisenkraft, Peter Goldreich, John Hopfield, Jon Kettenring, Geoffrey Lloyd, Donald Knuth, Carver Mead, David Middlebrook, Sterl Phinney, and Edwin Taylor.

感谢以下人员提出的许多宝贵建议和讨论： Shehu Abdussalam、Daniel Corbett、Dennis Freeman、Michael Godfrey、Hans Hagen、Jozef Hanc、Taco Hoekwater、Stephen Hou、Kayla Jacobs、Aditya Mahajan、Haynes Miller、Elisabeth Moyer、Hubert Pham、Benjamin Rapoport、Rahul Sarpeshkar、Madeleine Sheldon-Dante、Edwin Taylor、Tadashi Tokieda、Mark Warner 和 Joshua Zucker。

For many valuable suggestions and discussions: Shehu Abdussalam, Daniel Corbett, Dennis Freeman, Michael Godfrey, Hans Hagen, Jozef Hanc, Taco Hoekwater, Stephen Hou, Kayla Jacobs, Aditya Mahajan, Haynes Miller, Elisabeth Moyer, Hubert Pham, Benjamin Rapoport, Rahul Sarpeshkar, Madeleine Sheldon-Dante, Edwin Taylor, Tadashi Tokieda, Mark Warner, and Joshua Zucker.

有关写作过程的建议： Carver Mead 和 Hillary Rettig。

For advice on the process of writing: Carver Mead and Hillary Rettig.

有关书籍设计的建议： Yasuyo Iguchi。

For advice on the book design: Yasuyo Iguchi.

有关免费许可的建议： Daniel Ravicher 和 Richard Stallman。

For advice on free licensing: Daniel Ravicher and Richard Stallman.

对于用于计算的免费软件： Fredrik Johansson（mpmath）、Maxima 项目和 Python 社区。

For the free software used for calculations: Fredrik Johansson (mpmath), the Maxima project, and the Python community.

对于用于排版的自由软件： Hans Hagen 和 Taco Hoekwater (ConT _E Xt)；Han The Thanh (PDFT _E X)；Donald Knuth (T _E X)；John Hobby (MetaPost)；John Bowman、Andy Hammerlindl 和 Tom Prince (Asymptote)；Matt Mackall (Mercurial)；Richard Stallman (Emacs)；以及 Debian GNU/Linux 项目。

For the free software used for typesetting: Hans Hagen and Taco Hoekwater (ConT_EXt); Han The Thanh (PDFT_EX); Donald Knuth (T_EX); John Hobby (MetaPost); John Bowman, Andy Hammerlindl, and Tom Prince (Asymptote); Matt Mackall (Mercurial); Richard Stallman (Emacs); and the Debian GNU/Linux project.

感谢以下机构对我在科学和数学教学工作的支持：惠特克生物医学工程基金会、赫兹基金会、剑桥大学基督圣体学院院长和研究员、麻省理工学院教学与学习实验室和本科教育院长办公室；特别是罗杰·贝克、约翰·威廉姆斯和盖茨比慈善基金会的受托人。

For supporting my work in science and mathematics teaching: The Whitaker Foundation in Biomedical Engineering; the Hertz Foundation; the Master and Fellows of Corpus Christi College, Cambridge; the MIT Teaching and Learning Laboratory and the Office of the Dean for Undergraduate Education; and especially Roger Baker, John Williams, and the Trustees of the Gatsby Charitable Foundation.

一路顺风

Bon voyage

作为我们的第一个工具，让我们欢迎一位来自物理学和工程学的访客：维度分析方法。

As our first tool, let’s welcome a visitor from physics and engineering: the method of dimensional analysis.

1

1

方面

Dimensions

我们的第一个街头格斗工具是量纲分析，简称维度。为了展示其应用的多样性，我们以一个经济学例子来介绍该工具，并结合牛顿力学和积分学的例子进行进一步的阐释。

Our first street-fighting tool is dimensional analysis or, when abbreviated, dimensions. To show its diversity of application, the tool is introduced with an economics example and sharpened on examples from Newtonian mechanics and integral calculus.

1.1 经济学：跨国公司的力量

1.1 Economics: The power of multinational corporations

全球化的批评者经常做出如下比较[ 25 ]来证明跨国公司的权力过大：

Critics of globalization often make the following comparison [25] to prove the excessive power of multinational corporations:

尼日利亚是一个经济相对强劲的国家，其GDP（国内生产总值）为990亿美元。埃克森美孚的净资产为1190亿美元。“当跨国公司的净资产高于其经营所在国的GDP时，我们谈论的究竟是什么样的权力关系？”劳拉·莫罗西尼问道。

In Nigeria, a relatively economically strong country, the GDP [gross domestic product] is $99 billion. The net worth of Exxon is $119 billion. “When multinationals have a net worth higher than the GDP of the country in which they operate, what kind of power relationship are we talking about?” asks Laura Morosini.

在继续之前，请先探讨以下问题：

Before continuing, explore the following question:

埃克森美孚和尼日利亚之间的比较最严重的错误是什么？

What is the most egregious fault in the comparison between Exxon and Nigeria?

这个领域竞争激烈，但有一个缺陷显而易见。在解读GDP的含义后，这一点就显而易见了。GDP 990亿美元，其实就是每年990亿美元的货币流量。一年，也就是地球绕太阳公转一周的时间，是一种天文现象，被任意选择来衡量一种社会现象——货币流动。

The field is competitive, but one fault stands out. It becomes evident after unpacking the meaning of GDP. A GDP of $99 billion is shorthand for a monetary flow of $99 billion per year. A year, which is the time for the earth to travel around the sun, is an astronomical phenomenon that has been arbitrarily chosen for measuring a social phenomenon—namely, monetary flow.

假设经济学家选择十年作为衡量GDP的时间单位。那么尼日利亚的GDP（假设逐年保持稳定）每十年大约为1万亿美元，报告也为1万亿美元。现在，尼日利亚的GDP已经远远超过埃克森美孚，而埃克森美孚的资产仅为尼日利亚GDP的十分之一。为了得出相反的结论，假设以周作为衡量GDP的时间单位。尼日利亚的GDP变成了每周20亿美元，报告也为20亿美元。现在，孱弱的尼日利亚在强大的埃克森美孚面前束手无策，后者的规模是尼日利亚的50倍。

Suppose instead that economists had chosen the decade as the unit of time for measuring GDP. Then Nigeria’s GDP (assuming the flow remains steady from year to year) would be roughly $1 trillion per decade and be reported as $1 trillion. Now Nigeria towers over Exxon, whose puny assets are a mere one-tenth of Nigeria’s GDP. To deduce the opposite conclusion, suppose the week were the unit of time for measuring GDP. Nigeria’s GDP becomes $2 billion per week, reported as $2 billion. Now puny Nigeria stands helpless before the mighty Exxon, 50-fold larger than Nigeria.

有效的经济学论证不可能得出依赖于用于测量时间的天文现象的结论。其错误在于比较无法比较的量。净值是一个数量：它具有货币量纲，通常以美元为单位。然而，GDP是一个流量或速率：它具有单位时间的货币量纲，通常以美元/年为单位。（量纲是通用的，独立于测量体系，而单位则是该量纲在特定体系中的测量方式。）将净值与GDP进行比较，无异于将货币量与货币流量进行比较。由于它们的量纲不同，这种比较是一种范畴错误[ 39 ]，因此必然会产生无意义的结果。

A valid economic argument cannot reach a conclusion that depends on the astronomical phenomenon chosen to measure time. The mistake lies in comparing incomparable quantities. Net worth is an amount: It has dimensions of money and is typically measured in units of dollars. GDP, however, is a flow or rate: It has dimensions of money per time and typical units of dollars per year. (A dimension is general and independent of the system of measurement, whereas the unit is how that dimension is measured in a particular system.) Comparing net worth to GDP compares a monetary amount to a monetary flow. Because their dimensions differ, the comparison is a category mistake [39] and is therefore guaranteed to generate nonsense.

同样有缺陷的比较是单位时间长度（速度）与长度的比较：“我走1.5米/秒^——比300米高的纽约帝国大厦小得多。” 这完全是胡扯。为了得出相反但仍然胡扯的结论，可以用小时来衡量时间：“我走5400米/小时——比300米高的帝国大厦大得多。”

A similarly flawed comparison is length per time (speed) versus length: “I walk 1.5 ms⁻¹—much smaller than the Empire State building in New York, which is 300 m high.” It is nonsense. To produce the opposite but still nonsense conclusion, measure time in hours: “I walk 5400 m/hr—much larger than the Empire State building, which is 300 m high.”

我经常看到类似尼日利亚和埃克森美孚的例子，将企业实力与国家实力进行比较。我曾写信给一位作者，解释说我理解他的结论，但他的论点包含一个致命的维度错误。他回复说，我的观点很有意思，但用数字来比较国家实力，比他写的更有说服力，所以他保留了原样！

I often see comparisons of corporate and national power similar to our Nigeria–Exxon example. I once wrote to one author explaining that I sympathized with his conclusion but that his argument contained a fatal dimensional mistake. He replied that I had made an interesting point but that the numerical comparison showing the country’s weakness was stronger as he had written it, so he was leaving it unchanged!

一个维度上有效的比较是同类比较：要么将尼日利亚的GDP与埃克森美孚的收入进行比较，要么将埃克森美孚的净资产与尼日利亚的净资产进行比较。由于各国的净资产通常不以表格形式列出，而企业收入却很容易获得，因此可以尝试将埃克森美孚的年收入与尼日利亚的GDP进行比较。到2006年，埃克森美孚已成为埃克森美孚，年收入约为3500亿美元——几乎是尼日利亚2006年GDP（2000亿美元）的两倍。这种有效的比较比有缺陷的比较更有说服力，因此保留有缺陷的比较甚至不合时宜！

A dimensionally valid comparison would compare like with like: either Nigeria’s GDP with Exxon’s revenues, or Exxon’s net worth with Nigeria’s net worth. Because net worths of countries are not often tabulated, whereas corporate revenues are widely available, try comparing Exxon’s annual revenues with Nigeria’s GDP. By 2006, Exxon had become Exxon Mobil with annual revenues of roughly $350 billion—almost twice Nigeria’s 2006 GDP of $200 billion. This valid comparison is stronger than the flawed one, so retaining the flawed comparison was not even expedient!

被比较的量必须具有相同的量纲，这是进行有效比较的必要条件，但这还不够。一个代价高昂的例子是1999年的火星气候探测器（MCO），它撞上了火星表面，而不是滑入火星轨道。根据事故调查委员会（MIB）的说法，事故原因是英制和公制单位不匹配[ 26，第6页]：

That compared quantities must have identical dimensions is a necessary condition for making valid comparisons, but it is not sufficient. A costly illustration is the 1999 Mars Climate Orbiter (MCO), which crashed into the surface of Mars rather than slipping into orbit around it. The cause, according to the Mishap Investigation Board (MIB), was a mismatch between English and metric units [26, p. 6]:

MCO MIB 已确定 MCO 航天器坠毁的根本原因是未在用于轨迹模型的地面软件文件“小力”（Small Forces）的编码中使用公制单位。具体而言，在名为 SM_FORCES（小力）的软件应用程序代码中，使用的推进器性能数据采用英制单位而非公制单位。一个名为“角动量去饱和”（AMD）的文件包含 SM_FORCES 软件的输出数据。根据现有的软件接口文档，AMD 文件中的数据必须采用公制单位，而轨迹建模人员也假定数据符合要求，采用公制单位。

The MCO MIB has determined that the root cause for the loss of the MCO spacecraft was the failure to use metric units in the coding of a ground software file, Small Forces, used in trajectory models. Specifically, thruster performance data in English units instead of metric units was used in the software application code titled SM_FORCES (small forces). A file called Angular Momentum Desaturation (AMD) contained the output data from the SM_FORCES software. The data in the AMD file was required to be in metric units per existing software interface documentation, and the trajectory modelers assumed the data was provided in metric units per the requirements.

请务必注意您的尺寸和单位。

Make sure to mind your dimensions and units.

1.2 牛顿力学：自由落体

1.2 Newtonian mechanics: Free fall

维度不仅有助于推翻错误的论证，也有助于生成正确的论证。为此，问题中的量需要具有维度。以下是一个相反的例子，说明了哪些做法不该做：许多微积分教科书是这样引入一个经典问题的：

Dimensions are useful not just to debunk incorrect arguments but also to generate correct ones. To do so, the quantities in a problem need to have dimensions. As a contrary example showing what not to do, here is how many calculus textbooks introduce a classic problem in motion:

一个静止的球从 h英尺的高度落下，以 ν英尺/秒的速度撞击地面。假设重力加速度为 g英尺/秒²，忽略空气阻力，求 ν。

A ball initially at rest falls from a height of h feet and hits the ground at a speed of ν feet per second. Find ν assuming a gravitational acceleration of g feet per second squared and neglecting air resistance.

英尺或英尺每秒等单位以粗体突出显示，因为它们出现的频率太高，否则很容易被忽略，而它们的出现会造成严重的问题。由于高度为 h 英尺，变量 h 不包含高度单位：因此 h 是无量纲的。（如果 h 有量纲，那么问题就应该简单地表述为球从高度 h 落下；那么长度的量纲就属于 h。）类似地明确指定单位意味着变量 g 和 v 也是无量纲的。由于 g、h 和 ν 是无量纲的，因此将 ν 与由 g 和 h 导出的量进行比较，都是无量纲量之间的比较。因此，它在量纲上始终有效，量纲分析无法帮助我们猜测撞击速度。

The units such as feet or feet per second are highlighted in boldface because their inclusion is so frequent as to otherwise escape notice, and their inclusion creates a significant problem. Because the height is h feet, the variable h does not contain the units of height: h is therefore dimensionless. (For h to have dimensions, the problem would instead state simply that the ball falls from a height h; then the dimension of length would belong to h.) A similar explicit specification of units means that the variables g and v are also dimensionless. Because g, h, and ν are dimensionless, any comparison of ν with quantities derived from g and h is a comparison between dimensionless quantities. It is therefore always dimensionally valid, so dimensional analysis cannot help us guess the impact speed.

放弃维度这个宝贵的工具，就像是徒手搏斗，因此，我们必须在初始条件下求解以下微分方程：

Giving up the valuable tool of dimensions is like fighting with one hand tied behind our back. Thereby constrained, we must instead solve the following differential equation with initial conditions:

其中 y(t) 是球的高度，dy/dt 是球的速度，g 是重力加速度。

where y(t) is the ball’s height, dy/dt is the ball’s velocity, and g is the gravitational acceleration.

利用问题 1.3 中球的位置和速度的解，撞击速度是多少？

Using the solutions for the ball’s position and velocity in Problem 1.3, what is the impact speed?

当 y(t) = 0 时，球落地。因此撞击时间 t ₀为。撞击速度为 −gt ₀或 − 。因此，撞击速度（无符号速度）为。

When y(t) = 0, the ball meets the ground. Thus the impact time t₀ is . The impact velocity is −gt₀ or −. Therefore the impact speed (the unsigned velocity) is .

_{这种分析容易引发一些代数错误：求解 t 0}时忘记开平方，或者计算撞击速度时用除法而不是乘以 g。练习——换句话说，犯下并纠正许多错误——可以降低这些错误在简单问题中的发生率，但步骤繁多的复杂问题仍然是雷区。我们希望找到更不容易出错的方法。

This analysis invites several algebra mistakes: forgetting to take a square root when solving for t₀, or dividing rather than multiplying by g when finding the impact velocity. Practice—in other words, making and correcting many mistakes—reduces their prevalence in simple problems, but complex problems with many steps remain minefields. We would like less error-prone methods.

一种稳健的替代方案是量纲分析法。但这种方法要求ν、g和h中至少有一个量具有量纲。否则，所有候选撞击速度，无论多么荒谬，都等同于无量纲量，因此具有有效的量纲。

One robust alternative is the method of dimensional analysis. But this tool requires that at least one quantity among ν, g, and h have dimensions. Otherwise, every candidate impact speed, no matter how absurd, equates dimensionless quantities and therefore has valid dimensions.

因此，让我们重新表述自由落体问题，以便数量保留其维度：

Therefore, let’s restate the free-fall problem so that the quantities retain their dimensions:

一个静止的球从高度 h 落下，以速度 ν 撞击地面。假设重力加速度为 g，忽略空气阻力，求 ν。

A ball initially at rest falls from a height h and hits the ground at speed ν. Find ν assuming a gravitational acceleration g and neglecting air resistance.

首先，重述比原文措辞更简短、更清晰：

The restatement is, first, shorter and crisper than the original phrasing:

一个静止的球从 h 英尺的高度落下，以 ν 英尺/秒的速度落地。假设重力加速度为 g 英尺/秒²，忽略空气阻力，求 ν。

A ball initially at rest falls from a height of h feet and hits the ground at a speed of ν feet per second. Find ν assuming a gravitational acceleration of g feet per second squared and neglecting air resistance.

其次，重述更具通用性。它没有对单位制做出任何假设，因此即使长度单位是米、腕尺或弗隆，它仍然适用。最重要的是，重述给出了h、g和ν的量纲。它们的量纲几乎可以唯一地决定撞击速度——无需我们解微分方程。

Second, the restatement is more general. It makes no assumption about the system of units, so it is useful even if meters, cubits, or furlongs are the unit of length. Most importantly, the restatement gives dimensions to h, g, and ν. Their dimensions will almost uniquely determine the impact speed—without our needing to solve a differential equation.

高度 h 的量纲为长度，或简称为 L。重力加速度 g 的量纲为长度/时间平方，或 LT ⁻²，其中 T 表示时间维度。速度的量纲为 LT ⁻¹，因此 ν 是 g 和 h 的函数，其量纲为 LT ⁻¹。

The dimensions of height h are simply length or, for short, L. The dimensions of gravitational acceleration g are length per time squared or LT⁻², where T represents the dimension of time. A speed has dimensions of LT⁻¹, so ν is a function of g and h with dimensions of LT⁻¹.

g和h的哪种组合具有速度维度？

What combination of g and h has dimensions of speed?

这种组合具有速度维度。

The combination has dimensions of speed.

g和h的组合是否只具有速度维度？

Is the only combination of g and h with dimensions of speed?

为了确定是否是唯一的可能性，可以使用约束传播[ 43 ]。最强的约束是，g和h的组合（作为速度）应该具有逆时间（T ^-1）的维度。由于h不包含时间维度，因此它无法帮助构造T ^-1。因为g 包含 T ⁻²，则 T ⁻¹必定来自。第二个约束条件是 组合包含 L ¹。已经贡献了 L ^1/2，因此缺失的 L ^1/2必定来自这两个约束条件唯一地决定了 g 和 h 在撞击速度 ν 中的表现形式。

In order to decide whether is the only possibility, use constraint propagation [43]. The strongest constraint is that the combination of g and h, being a speed, should have dimensions of inverse time (T⁻¹). Because h contains no dimensions of time, it cannot help construct T⁻¹. Because g contains T⁻², the T⁻¹ must come from . The second constraint is that the combination contain L¹. The already contributes L^1/2, so the missing L^1/2 must come from The two constraints thereby determine uniquely how g and h appear in the impact speed ν.

然而，ν 的确切表达式并非唯一。它可能是、，或者通常为× 无量纲常数。乘以无量纲常数的惯用法经常出现，因此应该使用类似于等号的简洁符号：

The exact expression for ν is, however, not unique. It could be , , or, in general, × dimensionless constant. The idiom of multiplication by a dimensionless constant occurs frequently and deserves a compact notation akin to the equals sign:

包括这个∼符号在内，我们有几种相等的情况：

Including this ∼ notation, we have several species of equality:

精确的撞击速度为，因此维度结果包含整个函数依赖关系！它只缺少无量纲因子，而这些因素通常并不重要。在此示例中，高度可能从几厘米（跳蚤跳跃）到几米（猫从窗台上跳下来）不等。高度的 100 倍变化会导致撞击速度发生 10 倍变化。同样，重力加速度可能从 0.27 ms ⁻²（在小行星谷神星上）到 25 ms ⁻²（在木星上）不等。g 的 100 倍变化会导致撞击速度再次发生 10 倍变化。因此，撞击速度的许多变化不是来自无量纲因子，而是来自符号因子——这些符号因子是通过量纲分析精确计算出来的。

The exact impact speed is , so the dimensions result contains the entire functional dependence! It lacks only the dimensionless factor , and these factors are often unimportant. In this example, the height might vary from a few centimeters (a flea hopping) to a few meters (a cat jumping from a ledge). The factor-of-100 variation in height contributes a factor-of-10 variation in impact speed. Similarly, the gravitational acceleration might vary from 0.27 ms⁻² (on the asteroid Ceres) to 25 ms⁻² (on Jupiter). The factor-of-100 variation in g contributes another factor-of-10 variation in impact speed. Much variation in the impact speed, therefore, comes not from the dimensionless factor but rather from the symbolic factors—which are computed exactly by dimensional analysis.

此外，不计算精确答案也可能是一个优势。精确答案包含所有因数和项，这使得一些不太重要的信息（例如无量纲因子）能够掩盖诸如 之类的重要信息。正如威廉·詹姆斯所说：“智慧的艺术在于知道该忽略什么”[ 19，第22章]。

Furthermore, not calculating the exact answer can be an advantage. Exact answers have all factors and terms, permitting less important information, such as the dimensionless factor . to obscure important information such as . As William James advised, “The art of being wise is the art of knowing what to overlook” [19, Chapter 22].

1.3 猜测积分

1.3 Guessing integrals

自由落体运动的分析（第 1.2 节）表明，不将有量纲的量与其单位分开是有价值的。然而，如果这些量是无量纲的，例如以下高斯积分中的 5 和 x，情况又会怎样呢？

The analysis of free fall (Section 1.2) shows the value of not separating dimensioned quantities from their units. However, what if the quantities are dimensionless, such as the 5 and x in the following Gaussian integral:

或者，维度可能是未指定的——这在数学中很常见，因为它是一种通用语言。例如，概率论使用高斯积分

Alternatively, the dimensions might be unspecified—a common case in mathematics because it is a universal language. For example, probability theory uses the Gaussian integral

其中 x 可以是高度、探测器误差或其他参数。热物理学使用类似的积分

where x could be height, detector error, or much else. Thermal physics uses the similar integral

其中ν是分子速度。数学作为通用语言，研究它们的共同形式，而无需指定α和x的维度。缺乏具体性赋予了数学抽象的能力，但也使得量纲分析变得困难。

where ν is a molecular speed. Mathematics, as the common language, studies their common form without specifying the dimensions of α and x. The lack of specificity gives mathematics its power of abstraction, but it makes using dimensional analysis difficult.

如何应用维度分析而不失去数学抽象的好处？

How can dimensional analysis be applied without losing the benefits of mathematical abstraction?

答案是找到未指定维度的量，然后为它们分配一组一致的维度。为了说明这种方法，我们将其应用于广义定高斯积分

The answer is to find the quantities with unspecified dimensions and then to assign them a consistent set of dimensions. To illustrate the approach, let’s apply it to the general definite Gaussian integral

与其特定的近亲（α = 5，即积分）不同，一般形式没有指定 x 或 α 的维度——并且开放性提供了使用维度分析方法所需的自由。

Unlike its specific cousin with α = 5, which is the integral , the general form does not specify the dimensions of x or α—and that openness provides the freedom needed to use the method of dimensional analysis.

该方法要求任何方程的维度有效。因此，在下列方程中，左右两边的维度必须相同：

The method requires that any equation be dimensionally valid. Thus, in the following equation, the left and right sides must have identical dimensions:

右边是x的函数吗？是α的函数吗？它包含积分常数吗？

Is the right side a function of x? Is it a function of α? Does it contain a constant of integration?

左边除了 x 和 α 之外不包含任何符号量。但 x 是积分变量，且积分在一定范围内，所以积分时 x 消失（并且不出现积分常数）。因此，右边——那个“某物”——只是 α 的函数。用符号表示，

The left side contains no symbolic quantities other than x and α. But x is the integration variable and the integral is over a definite range, so x disappears upon integration (and no constant of integration appears). Therefore, the right side—the “something”—is a function only of α. In symbols,

函数 f 可能包含无量纲数，例如 2/3 或，但 α 是其唯一有量纲的输入。

The function f might include dimensionless numbers such as 2/3 or , but α is its only input with dimensions.

为了使方程在维度上有效，积分必须具有与 f(α) 相同的维度，并且 f(α) 的维度取决于 α 的维度。因此，维度分析程序包含以下三个步骤：

For the equation to be dimensionally valid, the integral must have the same dimensions as f(α), and the dimensions of f(α) depend on the dimensions of α. Accordingly, the dimensional-analysis procedure has the following three steps:

步骤 1. 为 α 分配尺寸（第 1.3.1 节）。

Step 1. Assign dimensions to α (Section 1.3.1).

步骤 2. 找到积分的维度（第 1.3.2 节）。

Step 2. Find the dimensions of the integral (Section 1.3.2).

步骤 3. 使用这些维度制作 f(α)（第 1.3.3 节）。

Step 3. Make an f(α) with those dimensions (Section 1.3.3).

1.3.1 为α分配维度

1.3.1 Assigning dimensions to α

参数 α 出现在指数中。指数指定将一个量乘以自身的次数。例如，这里是 2 ⁿ：

The parameter α appears in an exponent. An exponent specifies how many times to multiply a quantity by itself. For example, here is 2ⁿ:

“多少次”的概念是一个纯数字，因此指数是无量纲的。

The notion of “how many times” is a pure number, so an exponent is dimensionless.

因此，高斯积分中的指数 —αx ²是无量纲的。为方便起见，α 的维度记为 [α]，x 的维度记为 [x]。则

Hence the exponent —αx² in the Gaussian integral is dimensionless. For convenience, denote the dimensions of α by [α] and of x by [x]. Then

或者

or

这个结论很有用，但继续使用未指定但通用的维度需要大量的符号，而这些符号可能会掩盖推理。

This conclusion is useful, but continuing to use unspecified but general dimensions requires lots of notation, and the notation risks burying the reasoning.

最简单的替代方案是让 x 无量纲化。这样一来，α 和 f(α) 也就无量纲化了，因此 f(α) 的任何候选值在量纲上都是有效的，这样一来量纲分析就毫无意义了。最简单有效的替代方案是赋予 x 简单的量纲，例如长度。（如果你把 x 轴想象成平放在地板上，那么这个选择就很自然了。）那么 [α] = L ⁻²。

The simplest alternative is to make x dimensionless. That choice makes α and f(α) dimensionless, so any candidate for f(α) would be dimensionally valid, making dimensional analysis again useless. The simplest effective alternative is to give x simple dimensions—for example, length. (This choice is natural if you imagine the x axis lying on the floor.) Then [α] = L⁻².

1.3.2 积分的维度

1.3.2 Dimensions of the integral

赋值 [x] = L 和 [α] = L ⁻²决定了高斯积分的维数。以下是该积分：

The assignments [x] = L and [α] = L⁻² determine the dimensions of the Gaussian integral. Here is the integral again:

积分的维数取决于其三个部分的维数：积分符号 ∫、被积函数 e ^{−αx 2}和微分 dx。积分符号源于德语单词Summe的拉长字母 S ，Summe 意为“和”。在有效的和式中，所有项都具有相同的维数：维数的基本原理要求“苹果只能加在苹果上”。出于同样的原因，整个和式的维数与任何项都相同。因此，求和符号（以及积分符号）不影响维数：积分符号是无维数的。

The dimensions of an integral depend on the dimensions of its three pieces: the integral sign ∫, the integrand e^−αx2, and the differential dx. The integral sign originated as an elongated S for Summe, the German word for sum. In a valid sum, all terms have identical dimensions: The fundamental principle of dimensions requires that apples be added only to apples. For the same reason, the entire sum has the same dimensions as any term. Thus, the summation sign—and therefore the integration sign—do not affect dimensions: The integral sign is dimensionless.

由于积分符号无量纲，因此积分的量纲等于指数因子 e ^{−αx 2}的量纲乘以 dx 的量纲。指数函数尽管有 −αx ²这样的大指数，但实际上只是几个 e 的乘积。由于 e 是无量纲的，因此 e ^{−αx 2}也是无量纲的。

Because the integration sign is dimensionless, the dimensions of the integral are the dimensions of the exponential factor e^−αx2 multiplied by the dimensions of dx. The exponential, despite its fierce exponent −αx², is merely several copies of e multiplied together. Because e is dimensionless, so is e^−αx2.

dx的尺寸是多少？

What are the dimensions of dx?

要计算 dx 的维数，请遵循 Silvanus Thompson [ 45 , p. 1 ]的建议：将 d 读作“一点点”。那么 dx 就是“x 的一点点”。一点点长度仍然是长度，所以 dx 是长度。通常，dx 的维数与 x 相同。同样，d（∫ 的倒数）是无维数的。

To find the dimensions of dx, follow the advice of Silvanus Thompson [45, p. 1]: Read d as “a little bit of.” Then dx is “a little bit of x.” A little length is still a length, so dx is a length. In general, dx has the same dimensions as x. Equivalently, d—the inverse of ∫—is dimensionless.

组装各个部分，整个积分的长度尺寸为：

Assembling the pieces, the whole integral has dimensions of length:

1.3.3 制作具有正确维度的f(α)

1.3.3 Making an f(α) with correct dimensions

量纲分析的第三步也是最后一步是构造一个与积分维度相同的 f(α)。由于 α 的维度为 L ⁻²，因此将 α 转化为长度的唯一方法是构造 α ^−1/2。因此，

The third and final step in this dimensional analysis is to construct an f(α) with the same dimensions as the integral. Because the dimensions of α are L⁻², the only way to turn α into a length is to form α^−1/2. Therefore,

这个有用的结果仅缺少一个无量纲因子，并且无需任何积分即可获得。

This useful result, which lacks only a dimensionless factor, was obtained without any integration.

为了确定无量纲常数，设 α = 1 并评估

To determine the dimensionless constant, set α = 1 and evaluate

这个经典积分将在2.1 节中近似，猜测为。两个结果 f(1) =和 f(α) ∼ α ^−1/2要求 f(α) = ，从而得出

This classic integral will be approximated in Section 2.1 and guessed to be . The two results f(1) = and f(α) ∼ α^−1/2 require that f(α) = , which yields

我们经常记住无量纲常数，却忘记了α的幂。千万不要这样做。α因子通常比无量纲常数重要得多。方便的是，α因子是量纲分析可以计算出来的。

We often memorize the dimensionless constant but forget the power of α. Do not do that. The α factor is usually much more important than the dimensionless constant. Conveniently, the α factor is what dimensional analysis can compute.

1.4 总结和进一步的问题

1.4 Summary and further problems

不要把苹果和橘子混为一谈：等式或和式中的每个项必须具有相同的量纲！这个限制是一个强大的工具。它帮助我们在不积分的情况下求积分，并预测微分方程的解。这里有更多练习这个工具的问题。

Do not add apples to oranges: Every term in an equation or sum must have identical dimensions! This restriction is a powerful tool. It helps us to evaluate integrals without integrating and to predict the solutions of differential equations. Here are further problems to practice this tool.

2

2

简单案例

Easy cases

正确的解法适用于所有情况，包括简单情况。这条准则构成了第二个工具——简单情况法的基础。它将帮助我们猜测积分、推导体积，并求解复杂的微分方程。

A correct solution works in all cases, including the easy ones. This maxim underlies the second tool—the method of easy cases. It will help us guess integrals, deduce volumes, and solve exacting differential equations.

2.1 重温高斯积分

2.1 Gaussian integral revisited

作为第一个应用，让我们重新回顾一下第 1.3 节中的高斯积分，

As the first application, let’s revisit the Gaussian integral from Section 1.3,

是积分还是？

Is the integral or ?

正确的选择必须适用于所有。在此范围的端点 处，积分很容易计算。

The correct choice must work for all . At this range’s endpoints , the integral is easy to evaluate.

当

What is the integral when

像第一种简单情况一样，将 α 增加到 ∞ 时，即使 x 很小，− αx ²也会变得非常负。一个大的负数的指数很小，因此钟形曲线会变窄成一条细长条，其面积也会缩小到零。因此，当 α → ∞ 时，积分会缩小到零。这个结果反驳了以下选项，当 α = ∞ 时为无穷大；并且它支持选项，当 α = ∞ 时为零。

As the first easy case, increase α to ∞ Then − αx² becomes very negative, even when x is tiny. The exponential of a large negative number is tiny, so the bell curve narrows to a sliver, and its area shrinks to zero. Therefore, as α → ∞ the integral shrinks to zero. This result refutes the option , which is infinite when α = ∞; and it supports the option , which is zero when α = ∞.

当α = 0时积分是多少？

What is the integral when α = 0?

在 α = 0 的极端情况下，钟形曲线会变平为一条高度为单位的水平线。其面积在无穷大范围内积分后为无穷大。这一结果反驳了α = 0 时为零的选项；而支持了α = 0 时为无穷大的选项。因此，该选项既不通过两个简单情况测试，又通过了两个简单情况测试。

In the α = 0 extreme, the bell curve flattens into a horizontal line with unit height. Its area, integrated over the infinite range, is infinite. This result refutes the option, which is zero when α = 0; and it supports the option, which is infinity when α = 0. Thus the option fails both easy-cases tests, and the option passes both easy-cases tests.

如果只有这两个选项，我们会选择。但是，如果第三个选项是，你该如何在它和 之间做出选择？这两个选项都通过了简单情况测试；它们也具有相同的维度。选择看起来很困难。

If these two options were the only options, we would choose . However, if a third option were , how could you decide between it and ? Both options pass both easy-cases tests; they also have identical dimensions. The choice looks difficult.

要选择，请尝试第三种简单情况：α = 1。然后积分简化为

To choose, try a third easy case: α = 1. Then the integral simplifies to

这个经典积分可以用极坐标来求出闭式，但这种方法也需要一些技巧，而且应用范围有限（多元微积分的教科书会给出详细的说明）。一种不太优雅但更通用的方法是先数值计算积分，然后利用近似值来猜测闭式。

This classic integral can be evaluated in closed form by using polar coordinates, but that method also requires a trick with few other applications (textbooks on multivariable calculus give the gory details). A less elegant but more general approach is to evaluate the integral numerically and to use the approximate value to guess the closed form.

因此，将光滑曲线 e ^{−x 2}替换为包含 n 条线段的曲线。这种分段线性近似将面积转化为 n 个梯形的和。当 n 趋近于无穷大时，梯形的面积越来越接近光滑曲线下方的面积。

Therefore, replace the smooth curve e^−x2 with a curve having n line segments. This piecewise-linear approximation turns the area into a sum of n trapezoids. As n approaches infinity, the area of the trapezoids more and more closely approaches the area under the smooth curve.

该表给出了将曲线分成 n 条线段后，在 x = −10 ... 10 范围内的曲线下面积。面积最终稳定在一个值上，看起来很熟悉。它从 1.7 开始，这可能源于。然而，它继续下降到 1.77，这太大了，不可能是。幸运的是，π 略大于 3，因此面积可能收敛到。

The table gives the area under the curve in the range x = −10 . . . 10, after dividing the curve into n line segments. The areas settle onto a stable value, and it looks familiar. It begins with 1.7, which might arise from . However, it continues as 1.77, which is too large to be . Fortunately, π is slightly larger than 3, so the area might be converging to .

让我们通过将平方面积与π进行比较来检查：

Let’s check by comparing the squared area against π:

接近的匹配表明 α = 1 高斯积分确实是

The close match suggests that the α = 1 Gaussian integral is indeed

因此，一般的高斯积分

Therefore the general Gaussian integral

当 α = 1 时，必须减少到。它也必须在另外两个简单情况 α = 0 和 α = ∞ 下正确运行。

must reduce to when α = 1. It must also behave correctly in the other two easy cases α = 0 and α = ∞.

在三个选项中，只有α=0、1和∞三个检验通过。因此，

Among the three choices , , and , only passes all three tests α = 0, 1, and ∞. Therefore,

简单案例并非判断这些选择的唯一方法。例如，量纲分析也可以限制可能性（第 1.3 节）。它甚至会排除像通过所有三个简单案例测试那样的选择。然而，简单案例的设计本身就很简单。它们不需要我们为 x、α 和 dx 创造或推导维度（第 1.3 节的扩展分析）。与量纲分析不同，简单案例也可以排除像具有正确维度那样的选择。每种工具都有其优势。

Easy cases are not the only way to judge these choices. Dimensional analysis, for example, can also restrict the possibilities (Section 1.3). It even eliminates choices like that pass all three easy-cases tests. However, easy cases are, by design, simple. They do not require us to invent or deduce dimensions for x, α, and dx (the extensive analysis of Section 1.3). Easy cases, unlike dimensional analysis, can also eliminate choices like with correct dimensions. Each tool has its strengths.

2.2 平面几何：椭圆的面积

2.2 Plane geometry: The area of an ellipse

第二个简单应用来自平面几何：椭圆的面积。这个椭圆的长半轴为 a，短半轴为 b。对于它的面积 A，考虑以下候选：

The second application of easy cases is from plane geometry: the area of an ellipse. This ellipse has semimajor axis a and semiminor axis b. For its area A consider the following candidates:

每位候选人的优点和缺点是什么？

What are the merits or drawbacks of each candidate?

候选 A = ab ^{2 的}维数为 L ³，而面积的维数必定为 L ²。因此 ab ²必定是错误的。

The candidate A = ab² has dimensions of L³, whereas an area must have dimensions of L². Thus ab² must be wrong.

候选 A = a ² + b ²具有正确的尺寸（其余候选也一样），因此接下来的测试是关于半径 a 和 b 的简单情况。对于 a，最小极值 a = 0 会产生一个面积为零的无限细椭圆。然而，当 a = 0 时，候选 A = a ² + b ²会简化为 A = b ²而不是 0；因此 a ² + b ²未通过 a = 0 测试。

The candidate A = a² + b² has correct dimensions (as do the remaining candidates), so the next tests are the easy cases of the radii a and b. For a, the low extreme a = 0 produces an infinitesimally thin ellipse with zero area. However, when a = 0 the candidate A = a² + b² reduces to A = b² rather than to 0; so a² + b² fails the a = 0 test.

当 a = 0 时，候选 A = a ^{3 /b 正确地预测了面积为零。由于 a = 0 是一个有用的简单情况，并且轴标签 a 和 b 几乎可以互换，因此其对称对应项 b = 0 也应该是一个有用的简单情况。它也会产生一个面积为零的无限细椭圆；可惜的是，候选 a 3} /b 预测的面积为无限大，因此它无法通过 b = 0 的检验。剩下两个候选。

The candidate A = a³/b correctly predicts zero area when a = 0. Because a = 0 was a useful easy case, and the axis labels a and b are almost interchangeable, its symmetric counterpart b = 0 should also be a useful easy case. It too produces an infinitesimally thin ellipse with zero area; alas, the candidate a³/b predicts an infinite area, so it fails the b = 0 test. Two candidates remain.

候选 A = 2ab 看起来不错。当 a = 0 或 b = 0 时，实际面积和预测面积均为零，因此 A = 2ab 通过了两种简单情况的测试。进一步的测试需要第三种简单情况：a = b。此时椭圆变成了半径为 a 、面积为 πa² 的圆^。然而，候选 2ab 会简化为 A = 2a² ^，因此它未能通过 a = b 测试。

The candidate A = 2ab shows promise. When a = 0 or b = 0, the actual and predicted areas are zero, so A = 2ab passes both easy-cases tests. Further testing requires the third easy case: a = b. Then the ellipse becomes a circle with radius a and area πa². The candidate 2ab, however, reduces to A = 2a², so it fails the a = b test.

候选区域 A = πab 通过了所有三个测试：a = 0、b = 0 和 a = b。随着每次测试通过，我们对候选区域的信心不断增强；πab 确实是正确的区域（问题 2.4）。

The candidate A = πab passes all three tests: a = 0, b = 0, and a = b. With each passing test, our confidence in the candidate increases; and πab is indeed the correct area (Problem 2.4).

2.3 立体几何：截头金字塔的体积

2.3 Solid geometry: The volume of a truncated pyramid

高斯积分示例（第 2.1 节）和椭圆面积示例（第 2.2 节）展示了简单案例作为一种分析方法：用于检查公式是否正确。下一个复杂层次是将简单案例作为一种综合方法：用于构造公式。

The Gaussian-integral example (Section 2.1) and the ellipse-area example (Section 2.2) showed easy cases as a method of analysis: for checking whether formulas are correct. The next level of sophistication is to use easy cases as a method of synthesis: for constructing formulas.

举个例子，取一个方形底座的金字塔，用一把与底座平行的刀从其顶部切下一块。这个截头金字塔（称为平截头体）有一个方形底座和一个与底座平行的方形顶部。设h为其垂直高度，b为其底座边长，a为其顶部边长。

As an example, take a pyramid with a square base and slice a piece from its top using a knife parallel to the base. This truncated pyramid (called the frustum) has a square base and square top parallel to the base. Let h be its vertical height, b be the side length of its base, and a be the side length of its top.

截头金字塔的体积是多少？

What is the volume of the truncated pyramid?

让我们综合一下体积公式。它是三个长度 h、a 和 b 的函数。这些长度分为两类：高和底长。例如，将立体倒置会交换 a 和 b 的含义，但保留 h；并且没有任何简单的运算可以将高与 a 或 b 互换。因此，体积可能有两个因子，每个因子只包含一种或多种长度：

Let’s synthesize the formula for the volume. It is a function of the three lengths h, a, and b. These lengths split into two kinds: height and base lengths. For example, flipping the solid on its head interchanges the meanings of a and b but preserves h; and no simple operation interchanges height with a or b. Thus the volume probably has two factors, each containing a length or lengths of only one kind:

比例推理将确定 f；一点维度推理和大量简单案例推理将确定 g。

Proportional reasoning will determine f; a bit of dimensional reasoning and a lot of easy-cases reasoning will determine g.

f是多少：体积应如何取决于高度？

What is f : How should the volume depend on the height?

要计算 f，可以使用比例推理思维实验。将固体切成垂直的细条，每条都像一个石油钻芯；然后想象将 h 加倍。这种变化使每条细条的体积加倍，从而使总体积 V 加倍。因此 f ∼ h 且 V ∝ h：

To find f, use a proportional-reasoning thought experiment. Chop the solid into vertical slivers, each like an oil-drilling core; then imagine doubling h. This change doubles the volume of each sliver and therefore doubles the whole volume V. Thus f ∼ h and V ∝ h:

g是多少：体积应如何取决于a和b ？

What is g : How should the volume depend on a and b?

由于 V 的维数为 L ³，函数 g(a, b) 的维数为 L ²。该约束条件是维数分析所能给出的全部内容。为了合成 g ，还需要进一步的约束条件，这些约束条件可以通过简单案例法来提供。

Because V has dimensions of L³, the function g(a, b) has dimensions of L². That constraint is all that dimensional analysis can say. Further constraints are needed to synthesize g, and these constraints are provided by the method of easy cases.

2.3.1 简单案例

2.3.1 Easy cases

a和b的简单情况有哪些？

What are the easy cases of a and b?

最简单的情况是极端情况 a = 0（普通金字塔）。a 和 b 之间的对称性表明还有另外两种简单情况，即 a = b 和极端情况 b = 0。简单情况有三种：

The easiest case is the extreme case a = 0 (an ordinary pyramid). The symmetry between a and b suggests two further easy cases, namely a = b and the extreme case b = 0. The easy cases are then threefold:

当 a = 0 时，该立体是一个普通的金字塔，g 仅是底边长 b 的函数。由于 g 的维数为 L ^²，因此 g 的唯一可能性是 g ∼ b ^²；此外，V ∝ h；所以，V ∼ hb ^²。当 b = 0 时，该立体是一个 b = 0 金字塔的倒置版本，因此其体积为 V ∼ ha ^²。当 a = b 时，该立体是一个长方体，其体积为 V = ha ^²（或 hb ^²）。

When a = 0, the solid is an ordinary pyramid, and g is a function only of the base side length b. Because g has dimensions of L², the only possibility for g is g ∼ b²; in addition, V ∝ h; so, V ∼ hb². When b = 0, the solid is an upside-down version of the b = 0 pyramid and therefore has volume V ∼ ha². When a = b, the solid is a rectangular prism having volume V = ha² (or hb²).

是否存在满足三个简单情况约束的体积公式？

Is there a volume formula that satisfies the three easy-cases constraints?

a = 0 和 b = 0 的约束条件由对称和 V ∼ h(a ² + b ² ) 满足。如果缺失的无量纲常数为 1/2，使得 V = h(a ² + b ² )/2，则体积也满足 a = b 的约束条件，普通金字塔（a = 0）的体积为 hb ² /2。

The a = 0 and b = 0 constraints are satisfied by the symmetric sum V ∼ h(a² + b²). If the missing dimensionless constant is 1/2, making V = h(a² + b²)/2, then the volume also satisfies the a = b constraint, and the volume of an ordinary pyramid (a = 0) would be hb²/2.

当a = 0 时，预测V = hb ² /2正确吗？

When a = 0, is the prediction V = hb²/2 correct?

检验预测需要找到 V ∼ hb ²中的精确无量纲常数。这项任务看起来像一道微积分题：将金字塔切成薄片，然后将它们的体积相加（积分）。不过，一个简单的替代方法是再次应用简单的案例。

Testing the prediction requires finding the exact dimensionless constant in V ∼ hb². This task looks like a calculus problem: Slice a pyramid into thin horizontal sections and add (integrate) their volumes. However, a simple alternative is to apply easy cases again.

在我们解决了一个类似但更简单的问题之后，这个简单的情况就更容易构建了：求底边为 b、高为 h 的三角形的面积。面积满足 A ∼ hb，但无量纲常数是多少？要求它，选择 b 和 h 来构成一个简单的三角形：一个 h = b 的直角三角形。两个这样的三角形可以构成一个简单的形状：一个面积为 b² 的正方形^。因此，每个直角三角形的面积为 A = b² ^/ 2；无量纲常数为 1/2。现在将这个推理扩展到三维空间——找一个普通的金字塔（底边为正方形），将它与自身组合成一个简单的立体图形。

The easy case is easier to construct after we solve a similar but simpler problem: to find the area of a triangle with base b and height h. The area satisfies A ∼ hb, but what is the dimensionless constant? To find it, choose b and h to make an easy triangle: a right triangle with h = b. Two such triangles make an easy shape: a square with area b². Thus each right triangle has area A = b²/2; the dimensionless constant is 1/2. Now extend this reasoning to three dimensions—find an ordinary pyramid (with a square base) that combines with itself to make an easy solid.

什么是易固体？

What is the easy solid?

金字塔的方形底座暗示着一种方便的立体形状：也许每个底座都是立方体的一个面。这样，立方体就需要六个金字塔，它们的顶点在立方体的中心相交；因此，金字塔的长宽比为 h = b/2。为了简化计算，我们假设 b = 2 且 h = 1 满足此条件。

A convenient solid is suggested by the pyramid’s square base: Perhaps each base is one face of a cube. The cube then requires six pyramids whose tips meet in the center of the cube; thus the pyramids have the aspect ratio h = b/2. For numerical simplicity, let’s meet this condition with b = 2 and h = 1.

^{六个这样的金字塔组成一个体积为b 3} = 8的立方体，因此一个金字塔的体积为4/3。由于每个金字塔的体积V ∼ hb ²，且这些金字塔的hb ^{2 = 4，因此V ∼ hb 2}中的无量纲常数必定为1/3。因此，普通金字塔（a = 0的金字塔）的体积为V = hb ² /3。

Six such pyramids form a cube with volume b³ = 8, so the volume of one pyramid is 4/3. Because each pyramid has volume V ∼ hb², and hb² = 4 for these pyramids, the dimensionless constant in V ∼ hb² must be 1/3. The volume of an ordinary pyramid (a pyramid with a = 0) is therefore V = hb²/3.

前三个简单情况的测试预测 V = hb ² /2（当 a = 0 时），而另一个简单情况 h = b/2（当 a = 0 时）则显示 V = hb ² /3。这两种方法的预测结果相互矛盾。

The prediction from the first three easy-cases tests was V = hb²/2 (when a = 0), whereas the further easy case h = b/2 alongside a = 0 just showed that V = hb²/3. The two methods are making contradictory predictions.

如何解决这一矛盾？

How can this contradiction be resolved?

矛盾肯定是在某个推理步骤中潜入的。要找到罪魁祸首，请依次回顾每个步骤。V ∝ h 的论证看起来正确。三个简单情况的条件——当a = 0 时V ∼ hb ^{2 ，当 b = 0 时 V ∼ ha 2}，以及当 a = b 时 V = h(a ² + b ^{2 )/2——看起来也正确。错误在于从这些约束条件直接跳到对任意 a 或 b 都有 V ∼ h(a 2} + b ² ) 的预测。

The contradiction must have snuck in during one of the reasoning steps. To find the culprit, revisit each step in turn. The argument for V ∝ h looks correct. The three easy-case requirements—that V ∼ hb² when a = 0, that V ∼ ha² when b = 0, and that V = h(a² + b²)/2 when a = b—also look correct. The mistake was leaping from these constraints to the prediction V ∼ h(a² + b²) for any a or b.

相反，让我们尝试包含 ab 项的以下一般形式：

Instead let’s try the following general form that includes an ab term:

然后通过重新应用简单情况的要求来求解系数 α、β 和 γ。

Then solve for the coefficients α, β, and γ by reapplying the easy-cases requirements.

b = 0 检验以及 h = b/2 的简单情况（证明对于普通金字塔，V = hb ² /3）要求 α = 1/3。a = 0 检验同样要求 γ = 1/3。a = b 检验要求 α + β + γ = 1。因此 β = 1/3，瞧！

The b = 0 test along with the h = b/2 easy case, which showed that V = hb²/3 for an ordinary pyramid, require that α = 1/3. The a = 0 test similarly requires that γ = 1/3. And the a = b test requires that α + β + γ = 1. Therefore β = 1/3 and voilà,

这个公式是比例推理、量纲分析和简单案例方法的结果，是精确的（问题 2.9）！

This formula, the result of proportional reasoning, dimensional analysis, and the method of easy cases, is exact (Problem 2.9)!

2.4 流体力学：阻力

2.4 Fluid mechanics: Drag

前面的例子表明，简单情况可以检验并构造公式，但这些例子可以在没有简单情况的情况下完成（例如，使用微积分）。对于接下来的方程，由于流体力学中没有已知的精确解，因此简单情况和其他街头斗争工具几乎是取得进展的唯一途径。

The preceding examples showed that easy cases can check and construct formulas, but the examples can be done without easy cases (for example, with calculus). For the next equations, from fluid mechanics, no exact solutions are known in general, so easy cases and other street-fighting tools are almost the only way to make progress.

以下是流体力学的纳维-斯托克斯方程：

Here then are the Navier–Stokes equations of fluid mechanics:

其中，v是流体的速度（作为位置和时间的函数），ρ 是其密度，p 是压力，ν 是运动粘度。这些方程描述了各种各样的现象，包括飞行、龙卷风和急流。

where v is the velocity of the fluid (as a function of position and time), ρ is its density, p is the pressure, and ν is the kinematic viscosity. These equations describe an amazing variety of phenomena including flight, tornadoes, and river rapids.

我们的例子是下面这个关于阻力的家庭实验。复印此页并将其放大2倍；然后剪下以下两个模板：

Our example is the following home experiment on drag. Photocopy this page while magnifying it by a factor of 2; then cut out the following two templates:

用胶带将每个模板上的阴影部分粘在一起，做成一个圆锥体。两个圆锥体形状相同，但大圆锥体的高度和宽度是小圆锥体的两倍。

With each template, tape together the shaded areas to make a cone. The two resulting cones have the same shape, but the large cone has twice the height and width of the small cone.

当锥体朝下落下时，它们的终端速度（阻力平衡重量的速度）的近似比率是多少？

When the cones are dropped point downward, what is the approximate ratio of their terminal speeds (the speeds at which drag balances weight)?

纳维-斯托克斯方程包含了这个问题的答案。计算终端速度需要四个步骤。

The Navier–Stokes equations contain the answer to this question. Finding the terminal speed involves four steps.

步骤1.	施加边界条件。这些条件包括圆锥的运动以及不让流体进入纸张的要求。
第 2 步。	解该方程以及连续性方程= 0，以找到圆锥表面的压力和速度。
步骤3.	使用压力和速度来找到圆锥表面的压力和速度梯度；然后对产生的力进行积分以找到圆锥上的净力和扭矩。
步骤4.	利用净力和扭矩来计算圆锥的运动。这一步比较难，因为最终的运动必须与第一步假设的运动一致。如果不一致，就回到第一步，假设一个不同的运动，并祈祷到达这一步时运气更好。

遗憾的是，纳维-斯托克斯方程是耦合非线性偏微分方程。它们的解仅在非常简单的情况下才为人所知：例如，一个球体在粘性流体中以非常缓慢的速度运动，或者一个球体在零粘度流体中以任意速度运动。对于像柔性纸锥这样不规则、颤动形状的复杂流动，求解的希望微乎其微。

Unfortunately, the Navier–Stokes equations are coupled and nonlinear partial-differential equations. Their solutions are known only in very simple cases: for example, a sphere moving very slowly in a viscous fluid, or a sphere moving at any speed in a zero-viscosity fluid. There is little hope of solving for the complicated flow around an irregular, quivering shape such as a flexible paper cone.

2.4.1 使用尺寸

2.4.1 Using dimensions

由于无法直接求解纳维-斯托克斯方程，我们不妨采用量纲分析法和一些简单案例。直接方法是利用这些方法推导出终端速度本身。间接方法是推导出阻力与下落速度的关系，然后找到阻力与锥体重量平衡的速度。这种两步法简化了问题。它只引入了一个新量（阻力），但省去了两个量：重力加速度和锥体的质量。

Because a direct solution of the Navier–Stokes equations is out of the question, let’s use the methods of dimensional analysis and easy cases. A direct approach is to use them to deduce the terminal velocity itself. An indirect approach is to deduce the drag force as a function of fall speed and then to find the speed at which the drag balances the weight of the cones. This two-step approach simplifies the problem. It introduces only one new quantity (the drag force) but eliminates two quantities: the gravitational acceleration and the mass of the cone.

量纲原则是指有效方程中的所有项都具有相同的量纲。将其应用于阻力F，意味着在方程F = f（影响F的量）中，方程两边都具有力的量纲。因此，策略是先找到影响F的量，找到它们的量纲，然后将这些量合并成一个具有力的量纲的量。

The principle of dimensions is that all terms in a valid equation have identical dimensions. Applied to the drag force F, it means that in the equation F = f(quantities that affect F) both sides have dimensions of force. Therefore, the strategy is to find the quantities that affect F, find their dimensions, and then combine the quantities into a quantity with dimensions of force.

阻力取决于哪些量，它们的量纲是多少？

On what quantities does the drag depend, and what are their dimensions?

阻力取决于四个量：圆锥的两个参数和流体（空气）的两个参数。（关于ν的量纲，参见问题2.13。）

The drag force depends on four quantities: two parameters of the cone and two parameters of the fluid (air). (For the dimensions of ν, see Problem 2.13.)

四个参数ν、r、ρ和ν的任意组合是否都具有力的维度？

Do any combinations of the four parameters ν, r, ρ, and ν have dimensions of force?

下一步是将ν、r、ρ和ν组合成一个具有力的量纲的量。遗憾的是，可能性有很多——例如，

The next step is to combine ν, r, ρ, and ν into a quantity with dimensions of force. Unfortunately, the possibilities are numerous—for example,

或者和的乘积组合。这些不好看的乘积之和也是一种力，因此阻力 F 可能是、，甚至更糟。

or the product combinations and . Any sum of these ugly products is also a force, so the drag force F could be , , or much worse.

缩小可能性需要一种比简单地猜测具有正确维度的组合更复杂的方法。为了开发这种复杂的方法，需要回到维度的第一原理：方程中的所有项都具有相同的维度。该原理适用于任何关于阻力的陈述，例如

Narrowing the possibilities requires a method more sophisticated than simply guessing combinations with correct dimensions. To develop the sophisticated approach, return to the first principle of dimensions: All terms in an equation have identical dimensions. This principle applies to any statement about drag such as

其中，斑点 A、B 和 C 分别是 F、ν、r、ρ 和 ν 的函数。

where the blobs A, B, and C are functions of F, ν, r, ρ, and ν.

虽然这些斑点可能是极其复杂的函数，但它们的维度相同。因此，将每一项除以 A，得到公式

Although the blobs can be absurdly complex functions, they have identical dimensions. Therefore, dividing each term by A, which produces the equation

使每个项都无量纲。同样的方法可以将任何有效方程转化为无量纲方程。因此，任何描述世界的（真实）方程都可以写成无量纲形式。

makes each term dimensionless. The same method turns any valid equation into a dimensionless equation. Thus, any (true) equation describing the world can be written in a dimensionless form.

任何无量纲形式都可以由无量纲群构建：即由变量的无量纲乘积构建。由于任何描述世界的方程都可以写成无量纲形式，而任何无量纲形式都可以用无量纲群来写出，因此任何描述世界的方程都可以用无量纲群来写出。

Any dimensionless form can be built from dimensionless groups: from dimensionless products of the variables. Because any equation describing the world can be written in a dimensionless form, and any dimensionless form can be written using dimensionless groups, any equation describing the world can be written using dimensionless groups.

自由落体的例子（第 1.2 节）是否符合这一原则？

Is the free-fall example (Section 1.2) consistent with this principle?

在将此原理应用于复杂的阻力问题之前，先在自由落体这个简单的例子中尝试一下（第 1.2 节）。物体从高度 h 落下的精确撞击速度为 ν = ，其中 g 是重力加速度。这个结果实际上可以写成无量纲形式= ，而该形式本身只使用了无量纲组。新原理通过了初步检验。

Before applying this principle to the complicated problem of drag, try it in the simple example of free fall (Section 1.2). The exact impact speed of an object dropped from a height h is ν = where g is the gravitational acceleration. This result can indeed be written in the dimensionless form = , which itself uses only the dimensionless group . The new principle passes its first test.

这种公式的无量纲组分析，反过来就变成了一种综合方法。我们先来热身一下，综合一下冲击速度ν。首先，列出问题中的量；这里是ν、g和h。其次，将这些量组合成无量纲组。这里，所有无量纲组都可以从一个组构建出来。对于该组，我们选择ν2 ^/ gh（具体选择不影响结论）。那么唯一可能的无量纲命题是

This dimensionless-group analysis of formulas, when reversed, becomes a method of synthesis. Let’s warm up by synthesizing the impact speed ν. First, list the quantities in the problem; here, they are ν, g, and h. Second, combine these quantities into dimensionless groups. Here, all dimensionless groups can be constructed just from one group. For that group, let’s choose ν²/gh (the particular choice does not affect the conclusion). Then the only possible dimensionless statement is

（右侧是无量纲常数，因为那里没有第二个可用的组。）换句话说，ν ² /gh ∼ 1 或 ν ∼ 。

(The right side is a dimensionless constant because no second group is available to use there.) In other words, ν²/gh ∼ 1 or ν ∼ .

这一结果与1.2节中不太复杂的量纲分析的结果一致。事实上，如果只有一个无量纲群，两种分析方法都会得出相同的结论。然而，在诸如求阻力之类的难题中，不太复杂的方法无法以有用的形式提供其约束；这时，无量纲群方法就显得至关重要了。

This result reproduces the result of the less sophisticated dimensional analysis in Section 1.2. Indeed, with only one dimensionless group, either analysis leads to the same conclusion. However, in hard problems—for example, finding the drag force—the less sophisticated method does not provide its constraint in a useful form; then the method of dimensionless groups is essential.

对于阻力问题可以构建哪些无量纲群？

What dimensionless groups can be constructed for the drag problem?

一个无量纲群可以是 F/ρv ² r ²；另一个群可以是 rν/ν 。任何其他群都可以由这些群构造（问题 2.17 ），因此该问题由两个独立的无量纲群描述。因此，最一般的无量纲表述是

One dimensionless group could be F/ρv²r²; a second group could be rν/ν. Any other group can be constructed from these groups (Problem 2.17), so the problem is described by two independent dimensionless groups. The most general dimensionless statement is then

其中 f 是一个未知（但无量纲）的函数。

where f is a still-unknown (but dimensionless) function.

哪个无量纲组属于左侧？

Which dimensionless group belongs on the left side?

目标是合成一个 F 的公式，而 F 仅出现在第一组 F/ρv ² r ²中。考虑到这一限制，将第一组放在左侧，而不是将其包裹在仍然神秘的函数 f 中。基于此选择，关于阻力的最一般陈述是

The goal is to synthesize a formula for F, and F appears only in the first group F/ρv²r². With that constraint in mind, place the first group on the left side rather than wrapping it in the still-mysterious function f. With this choice, the most general statement about drag force is

圆锥体上（稳态）阻力的物理性质全部包含在无量纲函数 f 中。

The physics of the (steady-state) drag force on the cone is all contained in the dimensionless function f.

这个过程看似毫无意义，因为它产生的阻力取决于未知函数 f。但它大大提高了我们找到 f 的概率。最初的问题公式需要猜测 F = h(ν, r, ρ, ) 中的四变量函数 h，而量纲分析将问题简化为猜测一个只有一个变量（νr/ 比值）的函数。统计学家兼物理学家 Harold Jeffreys 曾雄辩地描述过这种简化的价值（引自 [ 34 , 第82页]）：

The procedure might seem pointless, having produced a drag force that depends on the unknown function f. But it has greatly improved our chances of finding f. The original problem formulation required guessing the four-variable function h in F = h(ν, r, ρ, ), whereas dimensional analysis reduced the problem to guessing a function of only one variable (the ratio νr/). The value of this simplification was eloquently described by the statistician and physicist Harold Jeffreys (quoted in [34, p. 82]):

一个好的单变量函数表可能需要一页纸；一个二变量函数表可能需要一卷书；一个三变量函数表可能需要一个书架；一个四变量函数表可能需要一个图书馆。

A good table of functions of one variable may require a page; that of a function of two variables a volume; that of a function of three variables a bookcase; and that of a function of four variables a library.

2.4.2 使用简单案例

2.4.2 Using easy cases

尽管情况有所改善，但我们找到答案的机会看起来并不大：即使是单变量阻力问题也没有精确解。但在简单的情形下，它可能有精确解。由于最简单的情况往往是极端情况，所以先看看极端情况。

Although improved, our chances do not look high: Even the one-variable drag problem has no exact solution. But it might have exact solutions in its easy cases. Because the easiest cases are often extreme cases, look first at the extreme cases.

极端情况有哪些？

Extreme cases of what?

未知函数 f 仅取决于 rν/ ，

The unknown function f depends on only rν/,

所以尝试 rν/ 的极端用法。然而，为了避免陷入盲目的符号推敲，首先要确定 rν/ 的含义。这个组合 rν/ ，通常用 Re 表示，是著名的雷诺数。（其物理解释需要使用集中技术，详见第 3.4.3 节。）

so try extremes of rν/ However, to avoid lapsing into mindless symbol pushing, first determine the meaning of rν/. This combination rν/ , often denoted Re, is the famous Reynolds number. (Its physical interpretation requires the technique of lumping and is explained in Section 3.4.3.)

雷诺数通过未知函数 f 影响阻力：

The Reynolds number affects the drag force via the unknown function f:

如果运气好的话，f 可以在雷诺数的极值处推导出来；如果运气更好的话，下落的锥体就是一个极端的例子。

With luck, f can be deduced at extremes of the Reynolds number; with further luck, the falling cones are an example of one extreme.

下落的锥体是雷诺数的极值吗？

Are the falling cones an extreme of the Reynolds number?

雷诺数取决于r、ν和。对于速度ν，日常经验表明，锥体下落的速度约为1 ms ⁻¹（例如，在2倍以内）。尺寸r约为0.1 m（同样在2倍以内）。空气的运动粘度约为10 ⁻⁵ m ² s ⁻¹。雷诺数为

The Reynolds number depends on r, ν, and . For the speed ν, everyday experience suggests that the cones fall at roughly 1 ms⁻¹ (within, say, a factor of 2). The size r is roughly 0.1 m (again within a factor of 2). And the kinematic viscosity of air is ∼ 10⁻⁵ m² s⁻¹. The Reynolds number is

它明显大于1，因此下落锥体是高雷诺数的极端情况。（对于低雷诺数的情况，请尝试问题2.27并参见[ 38 ]。）

It is significantly greater than 1, so the falling cones are an extreme case of high Reynolds number. (For low Reynolds number, try Problem 2.27 and see [38].)

高雷诺数极限可以通过多种方式达到。一种方法是将粘度减小到0，因为位于雷诺数的分母中。因此，在高雷诺数极限下，粘度从问题中消失，阻力不应依赖于粘度。这种推理包含一些微妙的谬误，但其结论基本正确。（澄清这些微妙之处需要数学两个世纪的进步，最终导致了奇异摄动和边界层理论的出现[ 12,46 ] 。）

The high-Reynolds-number limit can be reached many ways. One way is to shrink the viscosity to 0, because lives in the denominator of the Reynolds number. Therefore, in the limit of high Reynolds number, viscosity disappears from the problem and the drag force should not depend on viscosity. This reasoning contains several subtle untruths, yet its conclusion is mostly correct. (Clarifying the subtleties required two centuries of progress in mathematics, culminating in singular perturbations and the theory of boundary layers [12, 46].)

粘度仅通过雷诺数影响阻力：

Viscosity affects the drag force only through the Reynolds number:

为了使 F 与粘度无关，F 必须与雷诺数无关！这样，问题就只包含一个独立的无量纲群，F/ρv ² r ²，因此关于阻力的最一般表述是

To make F independent of viscosity, F must be independent of Reynolds number! The problem then contains only one independent dimensionless group, F/ρv²r², so the most general statement about drag is

阻力本身为 F ∼ ρv ² r ²。由于 r ²与圆锥的横截面积 A 成正比，因此阻力通常写为

The drag force itself is then F ∼ ρv²r². Because r² is proportional to the cone’s cross-sectional area A, the drag force is commonly written

虽然推导的是下落锥体，但只要雷诺数较高，结果也适用于任何物体。形状仅影响缺失的无量纲常数。对于球体，雷诺数大约为 1/4；对于垂直于其轴线运动的长圆柱体，雷诺数大约为 1/2；对于垂直于其表面运动的平板，雷诺数大约为 1。

Although the derivation was for falling cones, the result applies to any object as long as the Reynolds number is high. The shape affects only the missing dimensionless constant. For a sphere, it is roughly 1/4; for a long cylinder moving perpendicular to its axis, it is roughly 1/2; and for a flat plate moving perpendicular to its face, it is roughly 1.

2.4.3 终端速度

2.4.3 Terminal velocities

^{得出 F ∼ ρv 2} A的结果足以预测圆锥体的终端速度。终端速度意味着加速度为零，因此阻力必须与重量平衡。重量为 W = σ_纸A_纸g，其中 σ_纸是纸的面密度（单位面积质量），A_纸是模板切出四分之一截面后的面积。由于 A_纸与横截面积 A 相当，因此重量粗略地定义为

The result F ∼ ρv²A is enough to predict the terminal velocities of the cones. Terminal velocity means zero acceleration, so the drag force must balance the weight. The weight is W = σ_paperA_paperg, where σ_paper is the areal density of paper (mass per area) and A_paper is the area of the template after cutting out the quarter section. Because A_paper is comparable to the cross-sectional area A, the weight is roughly given by

所以，

Therefore,

面积除以终端速度

The area divides out and the terminal velocity becomes

所有由同一张纸构成且形状相同的圆锥体，无论其大小如何，都以相同的速度下落！

All cones constructed from the same paper and having the same shape, whatever their size, fall at the same speed!

为了验证这个预测，我制作了第21页描述的大、小锥体，双手各举一个，高过头顶，然后让它们落下。它们从2米高处落下，持续了大约2秒，落地间隔不到0.1秒。低成本实验和低成本理论都证明了这一点！

To test this prediction, I constructed the small and large cones described on page 21, held one in each hand above my head, and let them fall. Their 2 m fall lasted roughly 2 s, and they landed within 0.1 s of one another. Cheap experiment and cheap theory agree!

2.5 总结和进一步的问题

2.5 Summary and further problems

正确的解决方案适用于所有情况，包括简单情况。因此，请在简单情况下检查任何提出的公式，并通过构建通过所有简单情况测试的表达式来猜测公式。要应用和扩展这些想法，请尝试以下问题，并参阅Cipra的简明易懂的书籍[ 10 ]。

A correct solution works in all cases, including the easy ones. Therefore, check any proposed formula in the easy cases, and guess formulas by constructing expressions that pass all easy-cases tests. To apply and extend these ideas, try the following problems and see the concise and instructive book by Cipra [10].

3

3

集结

Lumping

一颗绕行行星六个月后会在哪里？为了预测它的新位置，我们不能简单地将六个月乘以行星的当前速度，因为它的速度是不断变化的。微积分正是为此而发明的。它的基本思想是将时间划分成速度恒定的微小区间，将每个微小区间乘以相应的速度，计算出一个微小的距离，然后将这些微小的距离相加。

Where will an orbiting planet be 6 months from now? To predict its new location, we cannot simply multiply the 6 months by the planet’s current velocity, for its velocity constantly varies. Such calculations are the reason that calculus was invented. Its fundamental idea is to divide the time into tiny intervals over which the velocity is constant, to multiply each tiny time by the corresponding velocity to compute a tiny distance, and then to add the tiny distances.

令人惊讶的是，即使区间宽度无穷小，因此数量无限，这种计算通常也能精确完成。然而，符号运算可能非常冗长，更糟糕的是，即使问题的微小变化也常常导致计算无法进行。例如，使用微积分方法，我们可以精确计算高斯函数 e ^{−x 2}在 x = 0 和 ∞ 之间的面积；然而，如果任一极限为零或无穷大以外的任何值，精确计算就变得不可能了。

Amazingly, this computation can often be done exactly, even when the intervals have infinitesimal width and are therefore infinite in number. However, the symbolic manipulations can be lengthy and, worse, are often rendered impossible by even small changes to the problem. Using calculus methods, for example, we can exactly calculate the area under the Gaussian e^−x2 between x = 0 and ∞; yet if either limit is any value except zero or infinity, an exact calculation becomes impossible.

相比之下，近似方法稳健：它们几乎总能给出合理的答案。准确度最低但稳健性最高的方法是集中法。与其将变化过程分解成许多小部分，不如将其组合或集中成一两个部分。我们将通过从人口统计学（第 3.1 节）到非线性微分方程（第 3.5 节）等示例来说明这种简单的近似方法及其优势。

In contrast, approximate methods are robust: They almost always provide a reasonable answer. And the least accurate but most robust method is lumping. Instead of dividing a changing process into many tiny pieces, group or lump it into one or two pieces. This simple approximation and its advantages are illustrated using examples ranging from demographics (Section 3.1) to nonlinear differential equations (Section 3.5).

3.1 估计人口：有多少婴儿？

3.1 Estimating populations: How many babies?

第一个例子是估算美国的婴儿数量。为了更明确起见，我们将婴儿（直到他或她2岁）称为婴儿。精确计算需要每个美国人的出生日期。美国人口普查局每十年收集一次此类信息或类似的信息。

The first example is to estimate the number of babies in the United States. For definiteness, call a child a baby until he or she turns 2 years old. An exact calculation requires the birth dates of every person in the United States. This, or closely similar, information is collected once every decade by the US Census Bureau.

为了近似这些海量数据，美国人口普查局[ 47 ]公布了各年龄段的人口数量。1991年的数据是一组位于一条曲线N(t)上的点，其中t代表年龄。那么

As an approximation to this voluminous data, the Census Bureau [47] publishes the number of people at each age. The data for 1991 is a set of points lying on a wiggly line N(t), where t is age. Then

这种方法存在几个问题。首先，它依赖于美国人口普查局的庞大资源，因此无法在荒岛上进行粗略计算。其次，它需要对一条没有解析形式的曲线进行积分，因此积分必须通过数值计算完成。第三，积分是针对特定问题的数据，而数学应该关注的是普遍性。简而言之，精确积分几乎无法提供任何洞见，转移价值也微乎其微。与其精确积分人口曲线，不如近似积分——将曲线集中到一个矩形中。

This method has several problems. First, it depends on the huge resources of the US Census Bureau, so it is not usable on a desert island for back-of-the-envelope calculations. Second, it requires integrating a curve with no analytic form, so the integration must be done numerically. Third, the integral is of data specific to this problem, whereas mathematics should be about generality. An exact integration, in short, provides little insight and has minimal transfer value. Instead of integrating the population curve exactly, approximate it—lump the curve into one rectangle.

这个矩形的高度和宽度是多少？

What are the height and width of this rectangle?

矩形的宽度代表一个时间，而与人口相关的一个合理时间是预期寿命。它大约是80年，所以假设每个人都在80岁生日那天突然去世，这样宽度就等于80年。矩形的高度可以根据矩形的面积计算出来，面积就是美国的人口——2008年恰好是3亿。因此，

The rectangle’s width is a time, and a plausible time related to populations is the life expectancy. It is roughly 80 years, so make 80 years the width by pretending that everyone dies abruptly on his or her 80th birthday. The rectangle’s height can be computed from the rectangle’s area, which is the US population—conveniently 300 million in 2008. Therefore,

为什么预期寿命从80岁下降到75岁？

Why did the life expectancy drop from 80 to 75 years?

捏造预期寿命简化了心算除法：75 很容易被 3 和 300 整除。其误差并不大于合并造成的误差，甚至可能抵消合并误差。以 75 年为宽度，高度约为 4 × 10 ⁶年⁻¹。

Fudging the life expectancy simplifies the mental division: 75 divides easily into 3 and 300. The inaccuracy is no larger than the error made by lumping, and it might even cancel the lumping error. Using 75 years as the width makes the height approximately 4 × 10⁶ yr⁻¹.

对 t = 0 . . . 2 年范围内的人口曲线进行积分就变成了乘法：

Integrating the population curve over the range t = 0 . . . 2 yr becomes just multiplication:

人口普查局的数据非常接近：7.980 × 10 ⁶。合并造成的误差抵消了将预期寿命捏造为 75 岁造成的误差！

The Census Bureau’s figure is very close: 7.980 × 10⁶. The error from lumping canceled the error from fudging the life expectancy to 75 years!

3.2 估算积分

3.2 Estimating integrals

美国人口曲线（第 3.1 节）难以积分，部分原因是它是未知的。但即使是众所周知的函数也可能难以积分。在这种情况下，两种集中方法特别有用：1/e 启发式（第 3.2.1 节）和半峰全宽 (FWHM) 启发式（第 3.2.2 节）。

The US population curve (Section 3.1) was difficult to integrate partly because it was unknown. But even well-known functions can be difficult to integrate. In such cases, two lumping methods are particularly useful: the 1/e heuristic (Section 3.2.1) and the full width at half maximum (FWHM) heuristic (Section 3.2.2).

3.2.1 1/e启发式

3.2.1 1/e heuristic

电子电路、大气压力和放射性衰变包含普遍存在的指数及其积分（此处以无量纲形式给出）

Electronic circuits, atmospheric pressure, and radioactive decay contain the ubiquitous exponential and its integral (given here in dimensionless form)

为了近似它的值，我们将 e ^−t曲线集中到一个矩形中。

To approximate its value, let’s lump the e^−t curve into one rectangle.

矩形的宽度和高度应该选择什么值？

What values should be chosen for the width and height of the rectangle?

^{矩形的合理高度是 e −t}的最大值，即 1。要选择其宽度，请使用显著变化作为标准（第 3.3.3 节中再次使用的方法）：选择 e ^−t的显著变化；然后找到产生此变化的宽度 Δt。在指数衰减中，一个简单而自然的显著变化是当 e ^−t成为 e 的一个因子，更接近其最终值（这里为 0，因为 t 趋于 ∞）。根据此标准，Δt = 1。则集中矩形具有单位面积——它是积分的精确值！

A reasonable height for the rectangle is the maximum of e^−t, namely 1. To choose its width, use significant change as the criterion (a method used again in Section 3.3.3): Choose a significant change in e^−t; then find the width Δt that produces this change. In an exponential decay, a simple and natural significant change is when e^−t becomes a factor of e closer to its final value (which is 0 here because t goes to ∞). With this criterion, Δt = 1. The lumping rectangle then has unit area—which is the exact value of the integral!

受到这个结果的鼓舞，让我们在困难的积分上尝试启发式方法

Encouraged by this result, let’s try the heuristic on the difficult integral

再次将面积集中到一个矩形中。它的高度是 e ^{−x ²}的最大值，即 1。它的宽度足以使 e ^{−x ²}下降 e 倍。这个下降发生在 x = ±1 处，因此宽度为 Δx = 2，面积为 1 × 2。精确面积约为1.77（见 2.1 节），因此集中计算的误差仅为 13%：对于如此简短的推导过程，精度极高。

Again lump the area into a single rectangle. Its height is the maximum of e^−x2 , which is 1. Its width is enough that e^−x2 falls by a factor of e. This drop happens at x = ±1, so the width is Δx = 2 and its area is 1 × 2. The exact area is ≈ 1.77 (Section 2.1), so lumping makes an error of only 13%: For such a short derivation, the accuracy is extremely high.

3.2.2 半峰全宽

3.2.2 Full width at half maximum

另一种合理的集中启发法出现在光谱学的早期。当光谱仪扫描一系列波长时，图表记录仪会绘制出分子吸收该波长辐射的强度。这条曲线包含许多峰，它们的位置和面积揭示了分子的结构（这对发展量子理论至关重要[ 14 ]）。但在数字图表记录仪出现的几十年前，这些峰的面积是如何计算的呢？

Another reasonable lumping heuristic arose in the early days of spectroscopy. As a spectroscope swept through a range of wavelengths, a chart recorder would plot how strongly a molecule absorbed radiation of that wavelength. This curve contains many peaks whose location and area reveal the structure of the molecule (and were essential in developing quantum theory [14]). But decades before digital chart recorders existed, how could the areas of the peaks be computed?

它们是通过将峰集中到一个矩形中来计算的，矩形的高度为峰高，宽度为半峰全宽 (FWHM)。1/e 启发式方法使用因子 e 作为显著变化，而 FWHM 启发式方法使用因子 2。

They were computed by lumping the peak into a rectangle whose height is the height of the peak and whose width is the full width at half maximum (FWHM). Where the 1/e heuristic uses a factor of e as the significant change, the FWHM heuristic uses a factor of 2.

在高斯积分上尝试这个方法。

Try this recipe on the Gaussian integral .

^{e −x 2}的最大高度为 1，因此半峰位于 x = ± 处，全宽为 2 。因此，集总矩形的面积 2 ≈ 1.665。精确面积约为1.77（见第 2.1 节）：FWHM 启发式方法的误差仅为 6%，大约是 1/e 启发式方法误差的一半。

The maximum height of e^−x2 is 1, so the half maxima are at x = ± and the full width is 2. The lumped rectangle therefore has area 2 ≈ 1.665. The exact area is ≈ 1.77 (Section 2.1): The FWHM heuristic makes an error of only 6%, which is roughly one-half the error of the 1/e heuristic.

3.2.3 斯特林近似

3.2.3 Stirling’s approximation

接下来，1/e 和 FWHM 集中启发式方法帮助我们近似普遍存在的阶乘函数 n!；该函数的用途广泛，从概率论到统计力学，再到算法分析。对于正整数，n! 定义为 n × (n − 1) × (n − 2) × · · · × 2 × 1。在这种离散形式下，它很难近似。然而，n! 的积分表示，

The 1/e and FWHM lumping heuristics next help us approximate the ubiquitous factorial function n!; this function’s uses range from probability theory to statistical mechanics and the analysis of algorithms. For positive integers, n! is defined as n × (n − 1) × (n − 2) × · · · × 2 × 1. In this discrete form, it is difficult to approximate. However, the integral representation for n!,

即使 n 不是正整数，也能提供定义——并且可以使用集中法来近似该积分。

provides a definition even when n is not a positive integer—and this integral can be approximated using lumping.

集中分析将生成几乎所有斯特林著名的近似公式

The lumping analysis will generate almost all of Stirling’s famous approximation formula

集中需要一个峰值，但被积函数t ⁿ e ^−t 有峰值吗？

Lumping requires a peak, but does the integrand tⁿe^−t have a peak?

要理解被积函数 t ⁿ e ^−t或 t ⁿ /e ^t，需要考察 t 的极端情况。当 t = 0 时，被积函数为 0。在另一个极端，即 t → ∞ 时，多项式因子 t ⁿ使乘积无穷大，而指数因子 e ^−t使其为零。谁会胜出呢？e ^t的泰勒级数包含 t 的所有幂（且系数为正），因此它是一个递增的无限次多项式。因此，当 t 趋于无穷大时，e ^t会超越任何多项式 t ⁿ，并使被积函数 t ⁿ /e ^t在 t → ∞ 的极端情况下等于 0。由于被积函数在两个极端都为零，因此在这两个极端之间必然有一个峰值。事实上，它只有一个峰值。（你能证明这一点吗？）

To understand the integrand tⁿe^−t or tⁿ/e^t, examine the extreme cases of t. When t = 0, the integrand is 0. In the opposite extreme, t → ∞, the polynomial factor tⁿ makes the product infinity while the exponential factor e^−t makes it zero. Who wins that struggle? The Taylor series for e^t contains every power of t (and with positive coefficients), so it is an increasing, infinite-degree polynomial. Therefore, as t goes to infinity, e^t outruns any polynomial tⁿ and makes the integrand tⁿ/e^t equal 0 in the t → ∞ extreme. Being zero at both extremes, the integrand must have a peak in between. In fact, it has exactly one peak. (Can you show that?)

随着 n 的增加，多项式因子 t ^{n 的}强度会增强，因此 t ⁿ会一直存在，直到 t 更高时，e ^{t才会超过它。因此，t n} /e ^t的峰值会随着 n 的增加而右移。该图证实了这一预测，并表明峰值出现在 t = n 时。让我们用微积分来验证一下，最大化 t ⁿ e ^−t，或者更简单地说，最大化其对数 f(t) = n ln t − t。在峰值处，函数的斜率为零。由于 df/dt = n/t−1，峰值出现在 t _peak = n 时，此时被积函数 t ⁿ e ^−t为 n ⁿ e ^−n——从而重现了斯特林公式中最大且最重要的因子。

Increasing n strengthens the polynomial factor tⁿ, so tⁿ survives until higher t before e^t outruns it. Therefore, the peak of tⁿ/e^t shifts right as n increases. The graph confirms this prediction and suggests that the peak occurs at t = n. Let’s check by using calculus to maximize tⁿe^−t or, more simply, to maximize its logarithm f(t) = n ln t − t. At a peak, a function has zero slope. Because df/dt = n/t−1, the peak occurs at t_peak = n, when the integrand tⁿe^−t is nⁿe⁻ⁿ—thus reproducing the largest and most important factor in Stirling’s formula.

什么是合理的集中矩形？

What is a reasonable lumping rectangle?

矩形的高度是峰值高度 n ⁿ e ⁻ⁿ。对于矩形的宽度，可以使用 1/e 或 FWHM 启发式方法。由于这两种启发式方法都需要近似 t ⁿ e ^−t，因此将其对数 f(t) 围绕 t = n 时的峰值以泰勒级数展开：

The rectangle’s height is the peak height nⁿe⁻ⁿ. For the rectangle’s width, use either the 1/e or the FWHM heuristics. Because both heuristic require approximating tⁿe^−t, expand its logarithm f(t) in a Taylor series around its peak at t = n:

泰勒展开式的第二项为零，因为 f(t) 在峰值处的斜率为零。第三项中，t = n 时的二阶导数 d ^² f/dt ^²为 −n/t ^²或 −1/n。因此，

The second term of the Taylor expansion vanishes because f(t) has zero slope at the peak. In the third term, the second derivative d²f/dt² at t = n is −n/t² or −1/n. Thus,

要将 t ⁿ e ^−t减小F 倍，需要将 f(t) 减小 ln F。这意味着。由于矩形的宽度为 2Δt，因此 n! 的集总面积估计值为

To decrease tⁿe^−t by a factor of F requires decreasing f(t) by ln F. This choice means . Because the rectangle’s width is 2Δt, the lumped-area estimate of n! is

相比之下，斯特林公式为 n! ≈ n ⁿ e ⁻ⁿ 。Lumping 几乎解释了所有因子。n ⁿ e ⁻ⁿ因子是矩形的高度，而因子是矩形的宽度。虽然确切的因子仍然未知（问题 3.9），但它的近似值在 13%（1/e 启发式）或 6%（FWHM 启发式）以内。

For comparison, Stirling’s formula is n! ≈ nⁿe⁻ⁿ Lumping has explained almost every factor. The nⁿe⁻ⁿ factor is the height of the rectangle, and the factor is from the width of the rectangle. Although the exact factor remains mysterious (Problem 3.9), it is approximated to within 13% (the 1/e heuristic) or 6% (the FWHM heuristic).

3.3 估计导数

3.3 Estimating derivatives

在前面的例子中，集中法有助于估计积分。由于积分和微分密切相关，集中法也提供了一种估算导数的方法。该方法始于对导数进行量纲观察。导数是微分的比值；例如，df/dx 是 df 与 dx 的比值。由于 d 是无量纲的（见第 1.3.2 节），因此 df/dx 的量纲就是 f/x 的量纲。这个有用而又令人惊奇的结论值得用一个熟悉的例子来检验：对高度 y 求时间 t 的微分，得到速度 dy/dt，其 LT ⁻¹的量纲实际上就是 y/t 的量纲。

In the preceding examples, lumping helped estimate integrals. Because integration and differentiation are closely related, lumping also provides a method for estimating derivatives. The method begins with a dimensional observation about derivatives. A derivative is a ratio of differentials; for example, df/dx is the ratio of df to dx. Because d is dimensionless (Section 1.3.2), the dimensions of df/dx are the dimensions of f/x. This useful, surprising conclusion is worth testing with a familiar example: Differentiating height y with respect to time t produces velocity dy/dt, whose dimensions of LT⁻¹ are indeed the dimensions of y/t.

3.3.1 正割近似

3.3.1 Secant approximation

由于 df/dx 和 f/x 具有相同的量纲，因此它们的大小可能相似：

As df/dx and f/x have identical dimensions, perhaps their magnitudes are similar:

从几何学上讲，导数 df/dx 是切线的斜率，而近似值 f/x 是割线的斜率。通过用割线代替曲线，我们得到了一个集中近似。

Geometrically, the derivative df/dx is the slope of the tangent line, whereas the approximation f/x is the slope of the secant line. By replacing the curve with the secant line, we make a lumping approximation.

让我们用一个简单的函数（例如 f(x) = x ^{2 ）}来测试这个近似值。好消息是，正割和正切的斜率仅相差 2 倍：

Let’s test the approximation on an easy function such as f(x) = x². Good news—the secant and tangent slopes differ only by a factor of 2:

f(x) = x ² + 100的正割近似值有多准确？

How accurate is the secant approximation for f(x) = x² + 100?

正割近似快速且实用，但可能会产生较大的误差。例如，当 f(x) = x ^{2 + 100 时，x = 1 处的正割和正切}斜率差异巨大。正切斜率 df/dx 为 2，而正割斜率 f(1)/1 为 101。这两个斜率的比值虽然无量纲，但却大得令人担忧。

The secant approximation is quick and useful but can make large errors. When f(x) = x² + 100, for example, the secant and tangent at x = 1 have dramatically different slopes. The tangent slope df/dx is 2, whereas the secant slope f(1)/1 is 101. The ratio of these two slopes, although dimensionless, is distressingly large.

替换导数 df/dx 的巨大差异在于

The large discrepancy in replacing the derivative df/dx, which is

割线斜率 f(x)/x 的近似值源于两个近似值。第一个近似值取 Δx = x 而不是 Δx = 0。则 df/dx ≈ (f(x) − f(0))/x。第一个近似值得出从 (0, f(0)) 到 (x, f(x)) 的直线的斜率。第二个近似值将 f(0) 替换为 0，得出 df/dx ≈ f/x；该比率是从 (0, 0) 到 (x, f(x)) 的割线的斜率。

with the secant slope f(x)/x is due to two approximations. The first approximation is to take Δx = x rather than Δx = 0. Then df/dx ≈ (f(x) − f(0))/x. This first approximation produces the slope of the line from (0, f(0)) to (x, f(x)). The second approximation replaces f(0) with 0, which produces df/dx ≈ f/x; that ratio is the slope of the secant from (0, 0) to (x, f(x)).

3.3.2 改进的割线近似

3.3.2 Improved secant approximation

第二个近似值是通过从 (0, f(0)) 而不是 (0, 0) 开始割线来解决的。

The second approximation is fixed by starting the secant at (0, f(0)) instead of (0, 0).

经过这种变化，当f(x) = x ^{2 + C}时，正割和正切的斜率是多少？

With that change, what are the secant and tangent slopes when f(x) = x² + C?

将从 (0, 0) 开始的割线称为原点割线；将新的割线称为 x = 0 割线。这样，无论常数 C 是多少，x = 0 割线的斜率始终是正切的一半。x = 0 割线近似值对垂直平移具有鲁棒性，即不受其影响。

Call the secant starting at (0, 0) the origin secant; call the new secant the x = 0 secant. Then the x = 0 secant always has one-half the slope of the tangent, no matter the constant C. The x = 0 secant approximation is robust against—is unaffected by—vertical translation.

x = 0正割近似对于水平平移的稳健性如何？

How robust is the x = 0 secant approximation against horizontal translation?

为了研究 x = 0 正割如何处理水平平移，将 f(x) = x ²向右平移 100 度，得到 f(x) = (x−100) ²。在抛物线的顶点 x = 100，x = 0 的割线，从 (0, 10 ⁴ ) 到 (100, 0)，斜率为 -100；然而，切线的斜率为零。因此，x = 0 的割线虽然比原点割线有所改进，但仍然会受到水平平移的影响。

To investigate how the x = 0 secant handles horizontal translation, translate f(x) = x² rightward by 100 to make f(x) = (x−100)². At the parabola’s vertex x = 100, the x = 0 secant, from (0, 10⁴) to (100, 0), has slope −100; however, the tangent has zero slope. Thus the x = 0 secant, although an improvement on the origin secant, is affected by horizontal translation.

3.3.3 显著变化近似值

3.3.3 Significant-change approximation

导数本身不受水平和垂直平移的影响，因此适当近似的导数可能具有平移不变性。近似导数为

The derivative itself is unaffected by horizontal and vertical translation, so a derivative suitably approximated might be translation invariant. An approximate derivative is

其中 Δx 不为零，但仍然很小。

where Δx is not zero but is still small.

Δx应该有多小？ Δx = 0.01够小吗？

How small should Δx be? Is Δx = 0.01 small enough?

选择 Δx = 0.01 有两个缺陷。首先，当 x 有维度时，它不起作用。如果 x 是长度，那么多长才足够小？选择 Δx = 1 毫米对于计算与太阳系相关的导数来说可能足够小，但对于计算与下落雾滴相关的导数来说可能太大。其次，没有固定的选择能够保持尺度不变。虽然 Δx = 0.01 在 f(x) = sin x 时可以产生精确的导数，但当 f(x) = sin 1000x（即简单地将 x 缩放到 1000x 的结果）时，它就失效了。

The choice Δx = 0.01 has two defects. First, it cannot work when x has dimensions. If x is a length, what length is small enough? Choosing Δx = 1 mm is probably small enough for computing derivatives related to the solar system, but is probably too large for computing derivatives related to falling fog droplets. Second, no fixed choice can be scale invariant. Although Δx = 0.01 produces accurate derivatives when f(x) = sin x, it fails when f(x) = sin 1000x, the result of simply rescaling x to 1000x.

这些问题建议尝试以下显著变化近似：

These problems suggest trying the following significant-change approximation:

因为这里的 Δx 由感兴趣点处的曲线属性定义，而不偏向特定的坐标值或 Δx 值，所以近似值是尺度和平移不变的。

Because the Δx here is defined by the properties of the curve at the point of interest, without favoring particular coordinate values or values of Δx, the approximation is scale and translation invariant.

为了说明这个近似值，我们尝试 f(x) = cos x，并用三个近似值估计 x = 3π/2 处的 df/dx：原点割线、x = 0 割线和显著变化近似。原点割线从 (0, 0) 到 (3π/2, 0)，因此斜率为零。它与精确斜率 1 的近似值很差。x = 0正割从 (0, 1) 到 (3π/2, 0)，因此它的斜率为 −2/3π，这比预测零斜率更糟糕，因为甚至符号都是错误的！

To illustrate this approximation, let’s try f(x) = cos x and estimate df/dx at x = 3π/2 with the three approximations: the origin secant, the x = 0 secant, and the significant-change approximation. The origin secant goes from (0, 0) to (3π/2, 0), so it has zero slope. It is a poor approximation to the exact slope of 1. The x = 0 secant goes from (0, 1) to (3π/2, 0), so it has a slope of −2/3π, which is worse than predicting zero slope because even the sign is wrong!

显著变化近似值可能更准确。f(x) = cos x 的显著变化是多少？由于余弦值变化了 2（从 −1 到 1），所以称 1/2 为 f(x) 的显著变化。当 x 从 3π/2（其中 f(x) = 0）变为 3π/2 + π/6（其中 f(x) = 1/2）时，就会发生这种变化。换句话说，Δx 等于 π/6。因此，近似导数为

The significant-change approximation might provide more accuracy. What is a significant change in f(x) = cos x? Because the cosine changes by 2 (from −1 to 1), call 1/2 a significant change in f(x). That change happens when x changes from 3π/2, where f(x) = 0, to 3π/2 + π/6, where f(x) = 1/2. In other words, Δx is π/6. The approximate derivative is therefore

该估计值约为 0.955 — 非常接近 1 的真实导数。

This estimate is approximately 0.955—amazingly close to the true derivative of 1.

3.4 分析微分方程：弹簧质量系统

3.4 Analyzing differential equations: The spring–mass system

估计导数将微分简化为除法（第 3.3 节）；从而将微分方程简化为代数方程。

Estimating derivatives reduces differentiation to division (Section 3.3); it thereby reduces differential equations to algebraic equations.

为了生成一个示例方程进行分析，将一个质量为 m 的块连接到一个弹簧常数（刚度）为 k 的理想弹簧上，将该块相对于平衡位置 x = 0 向右拉动 x ₀的距离，并在时间 t = 0 时释放它。块来回振荡，其位置 x 由理想弹簧微分方程描述

To produce an example equation to analyze, connect a block of mass m to an ideal spring with spring constant (stiffness) k, pull the block a distance x₀ to the right relative to the equilibrium position x = 0, and release it at time t = 0. The block oscillates back and forth, its position x described by the ideal-spring differential equation

让我们近似该方程并由此估计振荡频率。

Let’s approximate the equation and thereby estimate the oscillation frequency.

3.4.1 检查尺寸

3.4.1 Checking dimensions

看到任何方程式时，首先要检查它的量纲（第一章）。如果所有项的量纲不一致，则该方程式不值得求解——这将大大节省求解的精力。如果量纲一致，则检查结果会促使你思考这些项的含义；这种思考有助于为求解方程式和理解任何解做好准备。

Upon seeing any equation, first check its dimensions (Chapter 1). If all terms do not have identical dimensions, the equation is not worth solving—a great savings of effort. If the dimensions match, the check has prompted reflection on the meaning of the terms; this reflection helps prepare for solving the equation and for understanding any solution.

弹簧方程中两个项的量纲是多少？

What are the dimensions of the two terms in the spring equation?

首先看一下简单的第二项 kx。它源于胡克定律，该定律指出，理想弹簧施加的力为 kx，其中 x 是弹簧相对于其平衡长度的伸长量。因此，第二项 kx 是一个力。那么第一项也是力吗？

Look first at the simple second term kx. It arises from Hooke’s law, which says that an ideal spring exerts a force kx where x is the extension of the spring relative to its equilibrium length. Thus the second term kx is a force. Is the first term also a force?

第一项 m(d²x ^/ dt² ⁾包含二阶导数 d²x ^/ dt² ^，也就是我们熟悉的加速度。然而，许多微分方程包含我们不太熟悉的导数。流体力学的纳维-斯托克斯方程（第 2.4 节）

The first term m(d²x/dt²) contains the second derivative d²x/dt², which is familiar as an acceleration. Many differential equations, however, contain unfamiliar derivatives. The Navier–Stokes equations of fluid mechanics (Section 2.4),

包含两个奇怪的导数： ( v · Δ ) v和Δ ² v。这些项的量纲是多少？

contain two strange derivatives: (v·Δ)v and Δ²v. What are the dimensions of those terms?

为了练习稍后处理这些复杂的项，现在让我们手动计算 d ² x/dt ²的维度。由于 d ² x/dt ²包含两个 2 的指数，x 是长度，t 是时间，因此 d ² x/dt ²的维度可能为 L ² T ⁻²。

To practice for later handling such complicated terms, let’s now find the dimensions of d²x/dt² by hand. Because d²x/dt² contains two exponents of 2, and x is length and t is time, d²x/dt² might plausibly have dimensions of L²T⁻².

L ² T ⁻² 是正确的尺寸吗？

Are L²T⁻² the correct dimensions?

为了做出判断，我们可以使用1.3.2 节中的概念：微分符号 d 表示“一点点”。分子 d ^² x 表示 dx 的 d，是“x 的一点点”。因此，它是一个长度。分母 dt ^²可能表示 (dt) ^²或 d(t ^² )。[结果证明是 (dt) ^²。] 无论哪种情况，它的维度都是 T ^²。因此，二阶导数的维度为 LT ⁻²：

To decide, use the idea from Section 1.3.2 that the differential symbol d means “a little bit of.” The numerator d²x, meaning d of dx, is “a little bit of a little bit of x.” Thus, it is a length. The denominator dt² could plausibly mean (dt)² or d(t²). [It turns out to mean (dt)².] In either case, its dimensions are T². Therefore, the dimensions of the second derivative are LT⁻²:

这种组合就是加速度，因此弹簧方程的第一项 m(d ² x/dt ² ) 是质量乘以加速度——使其具有与 kx 项相同的维度。

This combination is an acceleration, so the spring equation’s first term m(d²x/dt²) is mass times acceleration—giving it the same dimensions as the kx term.

3.4.2 估计项的大小

3.4.2 Estimating the magnitudes of the terms

弹簧方程通过了量纲检验，因此值得分析以确定振动频率。方法是将每个项替换为其近似值。这些替换将把复杂的微分方程转化为简单的频率代数方程。

The spring equation passes the dimensions test, so it is worth analyzing to find the oscillation frequency. The method is to replace each term with its approximate magnitude. These replacements will turn a complicated differential equation into a simple algebraic equation for the frequency.

为了近似第一项 m(d ² x/dt ² )，使用显著变化近似法（第 3.3.3 节）来估计加速度 d ² x/dt ²的大小。

To approximate the first term m(d²x/dt²), use the significant-change approximation (Section 3.3.3) to estimate the magnitude of the acceleration d²x/dt².

要评估这个近似加速度，首先要确定一个显著的Δx，即什么能导致质量块位置发生显著变化。质量块在点 x = −x ₀和 x = +x ₀之间移动，因此位置的显著变化应该是峰峰值幅度 2x ₀的显著比例。最简单的选择是Δx = x ₀。

To evaluate this approximate acceleration, first decide on a significant Δx—on what constitutes a significant change in the mass’s position. The mass moves between the points x = −x₀ and x = +x₀, so a significant change in position should be a significant fraction of the peak-to-peak amplitude 2x₀. The simplest choice is Δx = x₀.

现在估计 Δt：即物块移动相当于 Δx 的距离所需的时间。这个时间称为系统的特征时间，与振荡周期 T 有关。在一个周期内，物块来回移动，移动的距离为 4x _{0 ，比 x 0}远得多。如果 Δt 为 T/4 或 T/2π，那么在 Δt 时间内物块移动的距离将与 x ₀相当。这些 Δt 的取值可以自然地解释为约为 1/ω，其中角频率 ω 与周期的关系定义：ω ≡ 2π/T。根据上述 Δx 和 Δt 的取值，m(d ² x/dt ² ) 项大约为 mx ₀ ω ²。

Now estimate Δt: the time for the block to move a distance comparable to Δx. This time—called the characteristic time of the system—is related to the oscillation period T. During one period, the mass moves back and forth and travels a distance 4x₀—much farther than x₀. If Δt were, say, T/4 or T/2π, then in the time Δt the mass would travel a distance comparable to x₀. Those choices for Δt have a natural interpretation as being approximately 1/ω, where the angular frequency ω is connected to the period by the definition ω ≡ 2π/T. With the preceding choices for Δx and Δt, the m(d²x/dt²) term is roughly mx₀ω².

“大致”是什么意思？

What does “is roughly” mean?

这句话的意思并非是 mx ₀ ω ²与 m(d ² x/dt ² ) 的差值在 2 倍以内，因为 m(d ² x/dt ² ) 会变化，而 mx ₀ /τ ²为常数。“大致”的意思是 m(d ² x/dt ² ) 的典型值或特征值（例如，其均方根值）与 mx ₀ ω ²相当。让我们将这个含义包含在旋转符号 ∼ 中。那么，典型值估计值可以写成

The phrase cannot mean that mx₀ω² and m(d²x/dt²) are within, say, a factor of 2, because m(d²x/dt²) varies and mx₀/τ² is constant. Rather, “is roughly” means that a typical or characteristic magnitude of m(d²x/dt²)—for example, its root-mean-square value—is comparable to mx₀ω². Let’s include this meaning within the twiddle notation ∼. Then the typical-magnitude estimate can be written

与“大致”的含义相同，即典型值是可比的，弹簧方程的第二项 kx 大致等于 kx _0。这两个项之和必须为零——这是弹簧方程的推论

With the same meaning of “is roughly”, namely that the typical magnitudes are comparable, the spring equation’s second term kx is roughly kx₀. The two terms must add to zero—a consequence of the spring equation

因此，这两个项的大小是可比较的：

Therefore, the magnitudes of the two terms are comparable:

振幅 x ₀除以 ω！当 x ₀消失后，频率 ω 和振荡周期 T = 2π/ω 与振幅无关。[此推论使用了几种近似值，但这个结论是精确的（问题 3.20）。] 近似的角频率 ω 为

The amplitude x₀ divides out! With x₀ gone, the frequency ω and oscillation period T = 2π/ω are independent of amplitude. [This reasoning uses several approximations, but this conclusion is exact (Problem 3.20).] The approximated angular frequency ω is then

为了进行比较，根据问题 3.22可知，弹簧微分方程的精确解为：

For comparison, the exact solution of the spring differential equation is, from Problem 3.22,

其中 ω 是近似的角频率也是精确的！

where ω is The approximated angular frequency is also exact!

3.4.3 雷诺数的含义

3.4.3 Meaning of the Reynolds number

作为集中的另一个例子——特别是显著变化近似——让我们分析一下第 2.4 节中介绍的纳维-斯托克斯方程，

As a further example of lumping—in particular, of the significant-change approximation—let’s analyze the Navier–Stokes equations introduced in Section 2.4,

并从中提取雷诺数rν/ν的物理意义。

and extract from them a physical meaning for the Reynolds number rν/ν.

为此，我们估计了惯性项 ( v · ∇ ) v和粘性项∇ ² v的典型大小。

To do so, we estimate the typical magnitude of the inertial term (v·∇)v and of the viscous term ∇²v.

惯性项的典型大小是多少？

What is the typical magnitude of the inertial term?

惯性项 ( v · ∇ ) ∇v包含空间导数∇v。根据显著变化近似（第 3.3.3 节），导数∇v大致等于

The inertial term (v·∇)∇v contains the spatial derivative ∇v. According to the significant-change approximation (Section 3.3.3), the derivative ∇v is roughly the ratio

远离圆锥体的流速（空气速度）几乎为零，而靠近圆锥体（以速度 ν 运动）时，其速度与 ν 相当。因此，ν 或 ν 的合理比例会引起流速的显著变化。这种速度变化发生在与圆锥体大小相当的距离上：在几个圆锥体长度之外，空气几乎感觉不到圆锥体下落。因此∇ν ∼ ν/r。惯性项 ( ν · ∇ ) ν包含v的第二个因子，因此 ( v · ∇ ) v大约等于 v ² /r。

The flow velocity (the velocity of the air) is nearly zero far from the cone and is comparable to ν near the cone (which is moving at speed ν). Therefore, ν, or a reasonable fraction of ν, constitutes a significant change in flow velocity. This speed change happens over a distance comparable to the size of the cone: Several cone lengths away, the air hardly knows about the falling cone. Thus ∇ν ∼ ν/r. The inertial term (ν·∇)ν contains a second factor of v, so (v·∇)v is roughly v²/r.

粘性项的典型大小是多少？

What is the typical magnitude of the viscous term?

粘性项 v ∇ ² v包含v的两个空间导数。由于每个空间导数都会对典型幅值贡献 1/r 的因子，因此 v∇ ² v大约等于 vν/r ²。惯性项与粘性项的比值大约等于 (ν ² /r)/(vν/r ² )。该比值简化为 rν/v——我们熟悉的无量纲雷诺数。

The viscous term v∇²v contains two spatial derivatives of v. Because each spatial derivative contributes a factor of 1/r to the typical magnitude, v∇²v is roughly vν/r². The ratio of the inertial term to the viscous term is then roughly (ν²/r)/(vν/r²). This ratio simplifies to rν/v—the familiar, dimensionless, Reynolds number.

因此，雷诺数衡量的是粘度的重要性。当 Re≫1 时，粘度项较小，粘度的影响可以忽略不计。它无法阻止邻近流体获得显著不同的速度，流动变为湍流。当 Re≫1 时，粘度项较大，粘度成为主要的物理效应。流动会像倒冷蜂蜜一样渗出。

Thus, the Reynolds number measures the importance of viscosity. When Re ≫ 1, the viscous term is small, and viscosity has a negligible effect. It cannot prevent nearby pieces of fluid from acquiring significantly different velocities, and the flow becomes turbulent. When Re 1, the viscous term is large, and viscosity is the dominant physical effect. The flow oozes, as when pouring cold honey.

3.5 预测单摆的周期

3.5 Predicting the period of a pendulum

集中不仅能将积分转化为乘法，还能将非线性微分方程转化为线性微分方程。我们以钟摆周期分析为例，几个世纪以来，钟摆一直是西方计时的基础。

Lumping not only turns integration into multiplication, it turns nonlinear into linear differential equations. Our example is the analysis of the period of a pendulum, for centuries the basis of Western timekeeping.

钟摆的周期与其振幅有何关系？

How does the period of a pendulum depend on its amplitude?

振幅 θ ₀是摆幅的最大角度；对于静止释放的无损耗摆，它也是释放角度。振幅的影响包含在摆微分方程的解中（方程的推导参见 [ 24 ]）：

The amplitude θ₀ is the maximum angle of the swing; for a lossless pendulum released from rest, it is also the angle of release. The effect of amplitude is contained in the solution to the pendulum differential equation (see [24] for the equation’s derivation):

分析将使用我们所有的工具：维度（第 3.5.2 节）、简单案例（第 3.5.1 节和第3.5.3节）和集中（第 3.5.4 节）。

The analysis will use all our tools: dimensions (Section 3.5.2), easy cases (Section 3.5.1 and Section 3.5.3), and lumping (Section 3.5.4).

3.5.1 小振幅：应用极端情况

3.5.1 Small amplitudes: Applying extreme cases

钟摆方程很难计算，因为它包含一个非线性因子 sin θ。幸运的是，在小振幅极端情况 θ → 0 下，该因子很容易计算。在该极限下，三角形的高 sin θ 几乎等于弧长 θ。因此，对于小角度，sin θ ≈ θ。

The pendulum equation is difficult because of its nonlinear factor sin θ. Fortunately, the factor is easy in the small-amplitude extreme case θ → 0. In that limit, the height of the triangle, which is sin θ, is almost exactly the arclength θ. Therefore, for small angles, sin θ ≈ θ.

在小振幅极端情况下，钟摆方程变为线性：

In the small-amplitude extreme, the pendulum equation becomes linear:

将此方程与弹簧质量方程（第 3.4 节）进行比较

Compare this equation to the spring–mass equation (Section 3.4)

方程中，x 类似于 θ，k/m 类似于 g/l。弹簧质量系统的频率为 ω = ，周期为 T = 2π/ω = 。对于单摆方程，相应的周期为

The equations correspond with x analogous to θ and k/m analogous to g/l. The frequency of the spring–mass system is ω = , and its period is T = 2π/ω = . For the pendulum equation, the corresponding period is

（本分析是对类比方法的预览，类比方法是第六章的主题。）

(This analysis is a preview of the method of analogy, which is the subject of Chapter 6.)

3.5.2 任意振幅：应用量纲分析

3.5.2 Arbitrary amplitudes: Applying dimensional analysis

如果振幅 θ ₀不再小，上述结果可能会发生变化。

The preceding results might change if the amplitude θ₀ is no longer small.

随着θ _{0 的} 增加，周期是增加、保持不变还是减少？

As θ₀ increases, does the period increase, remain constant, or decrease?

如果使用无量纲群（第 2.4.1 节）来表达，任何分析都会变得更加清晰。本问题涉及周期 T、长度 l、引力强度 g 和振幅 θ ₀。因此，T 可以属于无量纲群。由于角度是无量纲的，所以 θ ₀本身也是一个群。这两个群和 θ ₀相互独立，可以完整地描述该问题（问题 3.30）。

Any analysis becomes cleaner if expressed using dimensionless groups (Section 2.4.1). This problem involves the period T, length l, gravitational strength g, and amplitude θ₀. Therefore, T can belong to the dimensionless group . Because angles are dimensionless,θ₀ is itself a group. The two groups and θ₀ are independent and fully describe the problem (Problem 3.30).

一个有益的对比是理想的弹簧质量系统。周期 T、弹簧常数 k 和质量 m 可以构成无量纲组；但振幅 x ₀作为唯一包含长度的量，不能成为任何无量纲组的一部分（习题 3.20），因此不会影响弹簧质量系统的周期。相比之下，单摆的振幅θ ₀已经是一个无量纲群，因此它可以影响系统的周期。

An instructive contrast is the ideal spring–mass system. The period T, spring constant k, and mass m can form the dimensionless group ; but the amplitude x₀, as the only quantity containing a length, cannot be part of any dimensionless group (Problem 3.20) and cannot therefore affect the period of the spring–mass system. In contrast, the pendulum’s amplitude θ₀ is already a dimensionless group, so it can affect the period of the system.

两个无量纲群产生一般无量纲形式

Two dimensionless groups produce the general dimensionless form

所以

so

因为当θ ₀ = 0（小振幅极限）时，将2π分解出来以简化后续方程，并定义无量纲周期h如下：

Because when θ₀ = 0 (the small-amplitude limit), factor out the 2π to simplify the subsequent equations, and define a dimensionless period h as follows:

函数 h 包含了振幅如何影响单摆周期的所有信息。利用 h，关于周期的原始问题变成了：h 是振幅的增函数、常数函数还是减函数？下一节将解答这个问题。

The function h contains all information about how amplitude affects the period of a pendulum. Using h, the original question about the period becomes the following: Is h an increasing, constant, or decreasing function of amplitude? This question is answered in the following section.

3.5.3 大振幅：再次出现极端情况

3.5.3 Large amplitudes: Extreme cases again

为了猜测 h 作为振幅函数的一般行为，有用的线索来自于在两个振幅下评估 h。一个容易的振幅是零振幅的极值，其中 h(0) = 1。第二个容易的振幅是大振幅的另一个极值。

For guessing the general behavior of h as a function of amplitude, useful clues come from evaluating h at two amplitudes. One easy amplitude is the extreme of zero amplitude, where h(0) = 1. A second easy amplitude is the opposite extreme of large amplitudes.

在大振幅下，周期如何变化？作为这个问题的一部分，什么是大振幅？

How does the period behave at large amplitudes? As part of that question, what is a large amplitude?

一个有趣的大振幅是π/2，这意味着将摆锤从水平方向释放。然而，在π/2处，精确的h是以下可怕的表达式（问题3.31）：

An interesting large amplitude is π/2, which means releasing the pendulum from horizontal. However, at π/2 the exact h is the following awful expression (Problem 3.31):

这个积分小于、等于还是大于 1？谁知道呢？这个积分很可能没有封闭形式，需要数值计算（问题 3.32）。

Is this integral less than, equal to, or more than 1? Who knows? The integral is likely to have no closed form and to require numerical evaluation (Problem 3.32).

因为 θ ₀ = π/2 不是一个有用的极值，所以可以取更极端的值。试试 θ ₀ = π，这意味着将摆锤从垂直位置释放。然而，如果摆锤用绳子连接到支点，垂直释放将意味着摆锤垂直下落而不是振荡。这种新奇的行为既不包含在摆锤微分方程中，也无法用摆锤微分方程描述。

Because θ₀ = π/2 is not a helpful extreme, be even more extreme. Try θ₀ = π, which means releasing the pendulum bob from vertical. If the bob is connected to the pivot point by a string, however, a vertical release would mean that the bob falls straight down instead of oscillating. This novel behavior is neither included in nor described by the pendulum differential equation.

幸运的是，改进一个思想实验并不费力：用一根无质量的钢棒代替绳子。在θ ₀ = π时，摆锤完全平衡，永远倒挂着，因此T(π) = ∞，h(π) = ∞。因此，h(π) > 1，h(0) = 1。根据这些数据，最有可能的推测是h随振幅单调递增。虽然h可能先减小后增大，但对于这样一个纯粹的微分方程来说，这种曲折的行为会令人感到意外。（关于h在θ ₀ = π附近的行为，参见习题3.34）。

Fortunately, a thought experiment is cheap to improve: Replace the string with a massless steel rod. Balanced perfectly at θ₀ = π, the pendulum bob hangs upside down forever, so T(π) = ∞ and h(π) = ∞. Thus, h(π) > 1 and h(0) = 1. From these data, the most likely conjecture is that h increases monotonically with amplitude. Although h could first decrease and then increase, such twists and turns would be surprising behavior from such a clean differential equation. (For the behavior of h near θ₀ = π, see Problem 3.34).

3.5.4 中等振幅：应用集中法

3.5.4 Moderate amplitudes: Applying lumping

h单调递增的猜想是利用零和垂直振幅的极值推导出来的，因此它应该适用于中间振幅。在轻信这一说法之前，先回顾一下军控谈判中的一句谚语：“信任，但要核实。”

The conjecture that h increases monotonically was derived using the extremes of zero and vertical amplitude, so it should apply at intermediate amplitudes. Before taking that statement on faith, recall a proverb from arms-control negotiations: “Trust, but verify.”

在中等（小但不为零）振幅下，周期或其无量纲h是否会随着振幅的增加而增加？

At moderate (small but nonzero) amplitudes, does the period, or its dimensionless cousin h, increase with amplitude?

在零振幅极值时，sinθ接近于θ。该近似将非线性摆方程

In the zero-amplitude extreme, sin θ is close to θ. That approximation turned the nonlinear pendulum equation

进入线性理想弹簧方程——其中周期与振幅无关。

into the linear, ideal-spring equation—in which the period is independent of amplitude.

然而，在非零振幅时，θ 和 sin θ 不同，并且它们的差会影响周期。为了解释这个差异并预测周期，将 sin θ 分解为可处理因子 θ 和调整因子 f(θ)。所得方程为

At nonzero amplitude, however, θ and sin θ differ and their difference affects the period. To account for the difference and predict the period, split sin θ into the tractable factor θ and an adjustment factor f(θ). The resulting equation is

非常量 f(θ) 概括了摆方程的非线性。当 很小时，f(θ) ≈ 1：摆的行为类似于线性理想弹簧系统。但当 θ 较大时，f(θ) 会显著小于 1，使得理想弹簧近似值显著不准确。变化过程通常难以分析——例如，参见习题 3.31中那些糟糕的积分。为了解决这个问题，可以采用集中近似法，将变化的 f(θ) 替换为常数。

The nonconstant f(θ) encapsulates the nonlinearity of the pendulum equation. When is tiny, f(θ) ≈ 1: The pendulum behaves like a linear, ideal-spring system. But when θ is large, f(θ) falls significantly below 1, making the ideal-spring approximation significantly inaccurate. As is often the case, a changing process is difficult to analyze—for example, see the awful integrals in Problem 3.31. As a countermeasure, make a lumping approximation by replacing the changing f(θ) with a constant.

最简单的常数是 f(0)。则摆的微分方程变为

The simplest constant is f(0). Then the pendulum differential equation becomes

这个方程同样是理想弹簧方程。在这个近似中，周期与振幅无关，因此对于所有振幅，h = 1。为了确定未近似摆的周期如何依赖于振幅，f(θ) → f(0) 的集中近似会丢弃太多信息。

This equation is, again, the ideal-spring equation. In this approximation, period does not depend on amplitude, so h = 1 for all amplitudes. For determining how the period of an unapproximated pendulum depends on amplitude, the f(θ) → f(0) lumping approximation discards too much information.

因此，用另一个极值 f(θ ₀ ) 代替 f(θ)，则单摆方程变为

Therefore, replace f(θ) with the other extreme f(θ₀). Then the pendulum equation becomes

这个方程是线性的吗？它描述的是什么物理系统？

Is this equation linear? What physical system does it describe?

因为 f(θ ₀ ) 是一个常数，所以这个方程是线性的！它描述的是一个零振幅摆，摆位于一个重力 g _eff略弱于地球重力的行星上——如下图所示：

Because f(θ₀) is a constant, this equation is linear! It describes a zero-amplitude pendulum on a planet with gravity g_eff that is slightly weaker than earth gravity—as shown by the following slight regrouping:

因为零振幅摆的周期为，所以零振幅、低重力摆的周期为

Because the zero-amplitude pendulum has period , the zero-amplitude, low-gravity pendulum has period

使用无量纲周期 h 可以避免写出 2π、l 和 g 的因子，从而得到简单的预测

Using the dimensionless period h avoids writing the factors of 2π, l, and g, and it yields the simple prediction

在中等振幅下，近似值与精确的无量纲周期（深色曲线）非常接近。此外，它还预测了 h(π) = ∞，因此它与将钟摆从直立状态释放的思想实验（第 3.5.3 节）相符。

At moderate amplitudes the approximation closely follows the exact dimensionless period (dark curve). As a bonus, it also predicts h(π) = ∞, so it agrees with the thought experiment of releasing the pendulum from upright (Section 3.5.3).

10°振幅时的周期比零振幅时的周期大多少？

How much larger than the period at zero amplitude is the period at 10° amplitude?

10° 的振幅大约等于 0.17 弧度，这是一个中等角度，因此 h 的近似预测本身可以用泰勒级数精确地近似。sin 的泰勒级数起始于 θ−θ ³ /6，因此

A 10° amplitude is roughly 0.17 rad, a moderate angle, so the approximate prediction for h can itself accurately be approximated using a Taylor series. The Taylor series for sin begins θ−θ³/6, so

那么 h(θ ₀ )，大致等于 f(θ ₀ ) ^−1/2，变为

Then h(θ₀), which is roughly f(θ₀)^−1/2, becomes

另一个泰勒级数可得 (1 + x) ^−1/2 ≈ 1 − x/2（对于较小的 x）。因此，

Another Taylor series yields (1 + x)^−1/2 ≈ 1 − x/2 (for small x). Therefore,

恢复尺寸量即可得出周期本身。

Restoring the dimensioned quantities gives the period itself.

与零振幅时的周期相比，10°振幅会使周期略微增加，约为θ² ^/ ₀ /12≈0.0025，即0.25%。即使在中等振幅下，周期也几乎与振幅无关！

Compared to the period at zero amplitude, a 10° amplitude produces a fractional increase of roughly θ²₀/12 ≈ 0.0025 or 0.25%. Even at moderate amplitudes, the period is nearly independent of amplitude!

我们的集中近似是否低估或高估了周期？

Does our lumping approximation underestimate or overestimate the period?

_{集中近似法通过将 f(θ) 替换为 f(θ 0} )来简化摆的微分方程。等效地，它假设质量始终保持在运动的端点，即 |θ| = θ ₀。事实上，摆的大部分时间都处于中间位置，即 |θ| < θ ₀且 f(θ) > f(θ ₀ )。因此，平均 f 大于 f(θ ₀ )。由于 h 与 f 成反比（h = f ^−1/2），因此 f(θ) → f(θ ₀ ) 集中近似法会高估 h 和周期。

The lumping approximation simplified the pendulum differential equation by replacing f(θ) with f(θ₀). Equivalently, it assumed that the mass always remained at the endpoints of the motion where |θ| = θ₀. Instead, the pendulum spends much of its time at intermediate positions where |θ| < θ₀ and f(θ) > f(θ₀). Therefore, the average f is greater than f(θ₀). Because h is inversely related to f (h = f^−1/2), the f(θ) → f(θ₀) lumping approximation overestimates h and the period.

f(θ) → f(0) 的集中近似预测了 ，但低估了周期。因此，周期近似中 项的真实系数

The f(θ) → f(0) lumping approximation, which predicts , underestimates the period. Therefore, the true coefficient of the term in the period approximation

介于0和1/12之间。一个自然的猜测是，该系数位于这两个极值之间的中间值，即1/24。然而，钟摆在极值点（f(θ) = f(θ ₀ )）处花费的时间，比在平衡位置（f(θ) = f(0)）附近花费的时间要多。因此，真实系数可能更接近1/12（f(θ) → f(θ ₀ )近似值的预测值），而不是0。一个更合理的猜测可能是0到1/12之间三分之二的值，即1/18。

lies between 0 and 1/12. A natural guess is that the coefficient lies halfway between these extremes—namely, 1/24. However, the pendulum spends more time toward the extremes (where f(θ) = f(θ₀)) than it spends near the equilibrium position (where f(θ) = f(0)). Therefore, the true coefficient is probably closer to 1/12—the prediction of the f(θ) → f(θ₀) approximation—than it is to 0. An improved guess might be two-thirds of the way from 0 to 1/12, namely 1/18.

相比之下，单摆微分方程的完整逐次逼近解给出了以下周期 [ 13 , 33 ]：

In comparison, a full successive-approximation solution of the pendulum differential equation gives the following period [13, 33]:

我们根据经验猜测的 1/18 与真实系数 1/16 非常接近！

Our educated guess of 1/18 is very close to the true coefficient of 1/16!

3.6 总结和进一步的问题

3.6 Summary and further problems

集总法彻底颠覆了微积分。微积分通过将变化过程划分成越来越细的区间来分析它，而集总法则通过将变化过程合并成一个不变的过程来简化它。它把曲线变成直线，把复杂的积分变成乘法，把轻微的非线性微分方程变成线性微分方程。

Lumping turns calculus on its head. Whereas calculus analyzes a changing process by dividing it into ever finer intervals, lumping simplifies a changing process by combining it into one unchanging process. It turns curves into straight lines, difficult integrals into multiplication, and mildly nonlinear differential equations into linear differential equations.

……弯曲之处必变为正直，崎岖之处必变为平坦。（以赛亚书 40:4）

. . . the crooked shall be made straight, and the rough places plain. (Isaiah 40:4)

5

5

取出大部件

Taking out the big part

在几乎所有定量问题中，只要遵循“先做要事”这一谚语，分析就会变得简单。首先近似并理解最重要的影响因素——最重要的部分——然后完善你的分析和理解。这种逐次逼近或“去掉最重要的部分”的过程可以生成有意义、令人难忘且可用的表达式。以下示例介绍了低熵表达式的相关概念（第 5.2 节），并分析了心算乘法（第 5.1 节）、指数运算（第 5.3 节）、二次方程（第 5.4 节）以及一个困难的三角积分（第 5.5 节）。

In almost every quantitative problem, the analysis simplifies when you follow the proverbial advice of doing first things first. First approximate and understand the most important effect—the big part—then refine your analysis and understanding. This procedure of successive approximation or “taking out the big part” generates meaningful, memorable, and usable expressions. The following examples introduce the related idea of low-entropy expressions (Section 5.2) and analyze mental multiplication (Section 5.1), exponentiation (Section 5.3), quadratic equations (Section 5.4), and a difficult trigonometric integral (Section 5.5).

5.1 使用“一”和“少”进行乘法

5.1 Multiplication using one and few

第一个例子是一种适合粗略估算的心算乘法。具体计算的是数据 CD-ROM 的存储容量。数据 CD-ROM 的格式和存储容量与音乐 CD 相同，其容量可以通过以下三个因素的乘积来估算：

The first illustration is a method of mental multiplication suited to rough, back-of-the-envelope estimates. The particular calculation is the storage capacity of a data CD-ROM. A data CD-ROM has the same format and storage capacity as a music CD, whose capacity can be estimated as the product of three factors:

（在样本大小因素中，两个通道用于立体声。）

(In the sample-size factor, the two channels are for stereophonic sound.)

粗略计算使用诸如播放时间之类的粗略估计，而忽略了诸如用于错误检测和纠正的比特数等重要因素。在这个估算以及许多其他估算中，使用精确到小数点后3位的乘法就显得有些过度了。近似分析需要近似的计算方法。

Back-of-the-envelope calculations use rough estimates such as the playing time and neglect important factors such as the bits devoted to error detection and correction. In this and many other estimates, multiplication with 3 decimal places of accuracy would be overkill. An approximate analysis needs an approximate method of calculation.

数据容量在 2 倍以内是多少？

What is the data capacity to within a factor of 2?

单位（最大的部分！）是位（问题 5.3），三个数值因子贡献了 3600 × 4.4 × 10 ⁴ × 32。要估算乘积，将其分成大部分和校正部分。

The units (the biggest part!) are bits (Problem 5.3), and the three numerical factors contribute 3600 × 4.4 × 10⁴ × 32. To estimate the product, split it into a big part and a correction.

大数部分：粗略计算产品中最重要的因子通常来自 10 的幂，因此首先评估这个大数部分：3600 贡献 3 个 10 的幂，4.4 × 10 ^{4贡献 4 个，32 贡献 1 个。10 的 8 个幂产生了10}的因子。

The big part: The most important factor in a back-of-the-envelope product usually comes from the powers of 10, so evaluate this big part first: 3600 contributes three powers of 10, 4.4 × 10⁴ contributes four, and 32 contributes one. The eight powers of 10 produce a factor of 10⁸.

修正：去掉较大部分后，剩余部分是修正因子 3.6 × 4.4 × 3.2。此乘积也通过去掉较大部分简化。将每个因子四舍五入为三个选项中最接近的数字：1、few 或 10。我们发明的数字 few 介于 1 和 10 之间：它是 1 和 10 的几何平均值，因此 (few) ² = 10，few ≈ 3。在乘积 3.6×4.4×3.2 中，每个因子四舍五入为 few，因此 3.6×4.4×3.2 ≈ (few) ³，大约等于 30。

The correction: After taking out the big part, the remaining part is a correction factor of 3.6 × 4.4 × 3.2. This product too is simplified by taking out its big part. Round each factor to the closest number among three choices: 1, few, or 10. The invented number few lies midway between 1 and 10: It is the geometric mean of 1 and 10, so (few)² = 10 and few ≈ 3. In the product 3.6×4.4×3.2, each factor rounds to few, so 3.6×4.4×3.2 ≈ (few)³ or roughly 30.

单位、10 的幂和校正因子结合起来得出

The units, the powers of 10, and the correction factor combine to give

该估计值与精确乘积（问题 5.4 ）相差 2 倍以内，而精确乘积本身接近 5.6 × 10 ⁹位的实际容量。

This estimate is within a factor of 2 of the exact product (Problem 5.4), which is itself close to the actual capacity of 5.6 × 10⁹ bits.

5.2 分数变化和低熵表达式

5.2 Fractional changes and low-entropy expressions

使用“一或少数”法进行心算乘法很快。例如，3.15 × 7.21 很快就变成了“少数”× 10 ¹ ∼ 30，与精确乘积 22.7115 的误差在 50% 以内。为了得到更准确的估计，将 3.15 四舍五入为 3，将 7.21 四舍五入为 7。它们的乘积 21 的误差仅为 8%。为了进一步减少误差，可以将 3.15 × 7.21 分解成一个大数部分和一个加法修正部分。这种分解得到

Using the one-or-few method for mental multiplication is fast. For example, 3.15 × 7.21 quickly becomes few × 10¹ ∼ 30, which is within 50% of the exact product 22.7115. To get a more accurate estimate, round 3.15 to 3 and 7.21 to 7. Their product 21 is in error by only 8%. To reduce the error further, one could split 3.15 × 7.21 into a big part and an additive correction. This decomposition produces

这种方法虽然合理，但直接应用去掉大数部分会产生混乱的修正，难以记忆和理解。不过，稍加修改，去掉大数部分就能得到清晰直观的修正。此外，改进后的修正方法还引入了两个重要的“街头斗殴”思想：分数变化（第 5.2.1 节）和低熵表达式（第 5.2.2 节）。改进后的修正方法将作为众多用途中的第一个，帮助我们估算高速公路限速节省的能量（第 5.2.3 节）。

The approach is sound, but the literal application of taking out the big part produces a messy correction that is hard to remember and understand. Slightly modified, however, taking out the big part provides a clean and intuitive correction. As gravy, developing the improved correction introduces two important street-fighting ideas: fractional changes (Section 5.2.1) and low-entropy expressions (Section 5.2.2). The improved correction will then, as a first of many uses, help us estimate the energy saved by highway speed limits (Section 5.2.3).

5.2.1 分数变化

5.2.1 Fractional changes

加法校正的卫生替代方法是将产品分成大部分和乘法校正：

The hygienic alternative to an additive correction is to split the product into a big part and a multiplicative correction:

你能找到校正因子的图片吗？

Can you find a picture for the correction factor?

校正因子是一个宽为 1 + 0.05、高为 1 + 0.03 的矩形的面积。该矩形包含 (1 + 0.05) × (1 + 0.03) 展开式中每一项对应的子矩形。它们的面积之和约为 1 + 0.05 + 0.03，表示比大部分增加了 8%。大部分为 21，其中 8% 为 1.68，因此 3.15 × 7.21 = 22.68，与精确乘积的误差在 0.14% 以内。

The correction factor is the area of a rectangle with width 1 + 0.05 and height 1 + 0.03. The rectangle contains one subrectangle for each term in the expansion of (1 + 0.05) × (1 + 0.03). Their combined area of roughly 1 + 0.05 + 0.03 represents an 8% fractional increase over the big part. The big part is 21, and 8% of it is 1.68, so 3.15 × 7.21 = 22.68, which is within 0.14% of the exact product.

5.2.2 低熵表达式

5.2.2 Low-entropy expressions

3.15 × 7.21 的修正，作为绝对值或加性变化来说很复杂，但作为分数变化来说却很简单。这种对比具有普遍性。使用加性修正，双因子乘积变为

The correction to 3.15 × 7.21 was complicated as an absolute or additive change but simple as a fractional change. This contrast is general. Using the additive correction, a two-factor product becomes

当绝对变化 Δx 和 Δy 较小时（x Δx 和 y Δy），修正公式可简化为 xΔy + yΔx，但即便如此，它也很难记住，因为它包含许多看似合理但实际上不正确的替代项。例如，它可能包含 ΔxΔy、xΔx 或 yΔy 等项。合理替代方案的数量衡量了我们的直觉和现实之间的差距；差距越大，正确的结果就越难填补，我们也就越难记住正确的结果。

When the absolute changes Δx and Δy are small (x Δx and y Δy), the correction simplifies to xΔy + yΔx, but even so it is hard to remember because it has many plausible but incorrect alternatives. For example, it could plausibly contain terms such as ΔxΔy, xΔx, or yΔy. The extent of the plausible alternatives measures the gap between our intuition and reality; the larger the gap, the harder the correct result must work to fill it, and the harder we must work to remember the correct result.

此类差距是统计力学和信息论[20,21 ]的研究主题，它们将差距定义为合理替代方案数量的对数，并将对数值称为熵。对数并没有改变一个本质点：表达式在合理替代方案的数量上有所不同，而高熵表达式[ 28 ]（具有许多合理替代方案的表达式）难以记忆和理解。

Such gaps are the subject of statistical mechanics and information theory [20, 21], which define the gap as the logarithm of the number of plausible alternatives and call the logarithmic quantity the entropy. The logarithm does not alter the essential point that expressions differ in the number of plausible alternatives and that high-entropy expressions [28]—ones with many plausible alternatives—are hard to remember and understand.

相比之下，低熵表达式几乎不允许出现任何合理的替代方案，并会引发“是的！怎么可能不是这样呢？！”许多数学和科学进步都在于找到将高熵表达式转化为易于理解的低熵表达式的思维方式。

In contrast, a low-entropy expression allows few plausible alternatives, and elicits, “Yes! How could it be otherwise?!” Much mathematical and scientific progress consists of finding ways of thinking that turn high-entropy expressions into easy-to-understand, low-entropy expressions.

对乘积xy进行校正的低熵表达式是什么？

What is a low-entropy expression for the correction to the product xy?

由于乘法校正是无量纲的，因此其熵值自然低于加法校正：合理的无量纲表达式集比合理的表达式的完整集合小得多。

A multiplicative correction, being dimensionless, automatically has lower entropy than the additive correction: The set of plausible dimensionless expressions is much smaller than the full set of plausible expressions.

乘法校正为 (x + Δx)(y + Δy)/xy。顾名思义，该比率包含无偿熵。它构造两个有量纲的和 x + Δx 和 y + Δy，将它们相乘，最后将乘积除以 xy。虽然结果是无量纲的，但只有在最后一步才变为无量纲。一种更简洁的方法是通过直接创建无量纲量来对相关因子进行分组：

The multiplicative correction is (x + Δx)(y + Δy)/xy. As written, this ratio contains gratuitous entropy. It constructs two dimensioned sums x + Δx and y + Δy, multiplies them, and finally divides the product by xy. Although the result is dimensionless, it becomes so only in the last step. A cleaner method is to group related factors by making dimensionless quantities right away:

右侧仅由基本无量纲量 1 和有意义的无量纲比率构成：(Δx)/x 是 x 的分数变化，(Δy)/y 是 y 的分数变化。

The right side is built only from the fundamental dimensionless quantity 1 and from meaningful dimensionless ratios: (Δx)/x is the fractional change in x, and (Δy)/y is the fractional change in y.

无谓的熵来自于将 x + Δx、y + Δy、x 和 y 随意混合，它可以通过重组或分离来消除。在物理系统中，分离混合是很困难的。例如，试着将一滴食用色素混入一杯水中。问题是一杯水大约含有 10^ ²⁵个分子。幸运的是，大多数数学表达式的成分较少。我们通常可以重新组合并分离混合的成分，从而降低表达式的熵。

The gratuitous entropy came from mixing x + Δx, y + Δy, x, and y willy nilly, and it was removed by regrouping or unmixing. Unmixing is difficult with physical systems. Try, for example, to remove a drop of food coloring mixed into a glass of water. The problem is that a glass of water contains roughly 10²⁵ molecules. Fortunately, most mathematical expressions have fewer constituents. We can often regroup and unmix the mingled pieces and thereby reduce the entropy of the expression.

低熵校正因子产生低熵分数变化：

A low-entropy correction factor produces a low-entropy fractional change:

其中 Δ(xy)/xy 是从 xy 到 (x + Δx)(y + Δy) 的分数变化。最右边的项是两个小分数的乘积，因此与前两个项相比较小。如果没有这个小的二次项，

where Δ(xy)/xy is the fractional change from xy to (x + Δx)(y + Δy). The rightmost term is the product of two small fractions, so it is small compared to the preceding two terms. Without this small, quadratic term,

只需添加微小的分数变化即可！

Small fractional changes simply add!

这个分数变化规则比相应的近似规则（绝对变化为 xΔy + yΔx）简单得多。简单性表明熵值较低；事实上，该规则唯一可行的替代方案是分数变化成倍增加的可能性。而这个猜想不太可能成立：当 Δy = 0 时，它预测无论 Δx 的值如何，Δ(xy) = 0（该预测也在问题 5.12中进行了探讨）。

This fractional-change rule is far simpler than the corresponding approximate rule that the absolute change is xΔy + yΔx. Simplicity indicates low entropy; indeed, the only plausible alternative to the proposed rule is the possibility that fractional changes multiply. And this conjecture is not likely: When Δy = 0, it predicts that Δ(xy) = 0 no matter the value of Δx (this prediction is explored also in Problem 5.12).

5.2.3 平方

5.2.3 Squaring

在分析工程世界和自然世界时，一个常见的运算是平方——乘法的一种特殊情况。长度的平方等于面积，速度的平方与大多数物体所受的阻力成正比（见第 2.4 节）：

In analyzing the engineered and natural worlds, a common operation is squaring—a special case of multiplication. Squared lengths are areas, and squared speeds are proportional to the drag on most objects (Section 2.4):

其中，ν 为物体速度，A 为横截面积，ρ 为流体密度。因此，以高速公路速度行驶 d 的距离会消耗能量 E = F _d d ∼ ρAv ² d。因此，降低行驶速度可以降低能耗。在 20 世纪 70 年代油价快速上涨的背景下，这种可能性对西方国家变得尤为重要（参见 [ 7 ] 的分析）。因此，美国将高速公路限速设为 55 英里/小时（90 公里/小时）。

where ν is the speed of the object, A is its cross-sectional area, and ρ is the density of the fluid. As a consequence, driving at highway speeds for a distance d consumes an energy E = F_dd ∼ ρAv²d. Energy consumption can therefore be reduced by driving more slowly. This possibility became important to Western countries in the 1970s when oil prices rose rapidly (see [7] for an analysis). As a result, the United States instituted a highway speed limit of 55 mph (90 kph).

如果以55英里/小时而不是65英里/小时的速度行驶，汽油消耗量会下降多少？

By what fraction does gasoline consumption fall due to driving 55 mph instead of 65 mph?

降低限速可以通过降低阻力ρAν ²和缩短行驶距离d来降低油耗：人们更多地根据时间而非距离来衡量和调节通勤。但寻找新住所或新工作是一个缓​​慢的过程。因此，先分析重要事项——在初步分析中假设行驶距离d保持不变（然后尝试解答问题5.14）。

A lower speed limit reduces gasoline consumption by reducing the drag force ρAν² and by reducing the driving distance d: People measure and regulate their commuting more by time than by distance. But finding a new home or job is a slow process. Therefore, analyze first things first—assume for this initial analysis that the driving distance d stays fixed (then try Problem 5.14).

^{根据该假设，E 与 ν 2}成正比，并且

With that assumption, E is proportional to ν², and

从65英里/小时加速到55英里/小时，ν大约下降了15%，因此能耗下降了大约30%。高速公路行驶消耗了机动车消耗的石油总量的很大一部分，而在美国，机动车消耗的石油占总石油消耗的很大一部分。因此，30%的降幅大幅降低了美国的石油总消耗量。

Going from 65 mph to 55 mph is roughly a 15% drop in ν, so the energy consumption drops by roughly 30%. Highway driving uses a significant fraction of the oil consumed by motor vehicles, which in the United States consume a significant fraction of all oil consumed. Thus the 30% drop substantially reduced total US oil consumption.

5.3 具有一般指数的分数变化

5.3 Fractional changes with general exponents

^{x 2}变化的分数变化近似值（第 5.2.3 节）和 x ³变化的分数变化近似值（问题 5.13 ）是 x ⁿ近似值的特例

The fractional-change approximations for changes in x² (Section 5.2.3) and in x³ (Problem 5.13) are special cases of the approximation for xⁿ

这条规则提供了一种心算除法（第 5.3.1 节）、估算平方根（第 5.3.2 节）以及判断季节变化的共同解释（第 5.3.3 节）的方法。该规则仅要求分数变化较小且指数 n 不能太大（第 5.3.4 节）。

This rule offers a method for mental division (Section 5.3.1), for estimating square roots (Section 5.3.2), and for judging a common explanation for the seasons (Section 5.3.3). The rule requires only that the fractional change be small and that the exponent n not be too large (Section 5.3.4).

5.3.1 快速心理除法

5.3.1 Rapid mental division

n = −1 的特殊情况提供了快速心算除法的方法。例如，我们估算 1/13。将其重写为 (x + Δx) ⁻¹，其中 x = 10，Δx = 3。大的部分是 x ^{−1 = 0.1。由于 (Δx)/x = 30%，因此对 x −1 的}小数修正大约为 −30%。结果为 0.07。

The special case n = −1 provides the method for rapid mental division. As an example, let’s estimate 1/13. Rewrite it as (x + Δx)⁻¹ with x = 10 and Δx = 3. The big part is x⁻¹ = 0.1. Because (Δx)/x = 30%, the fractional correction to x⁻¹ is roughly −30%. The result is 0.07.

其中“−30%”符号表示“将前一个对象减少 30%”，是因子 1 − 0.3 的有用简写。

where the “−30%” notation, meaning “decrease the previous object by 30%,” is a useful shorthand for a factor of 1 − 0.3.

估计的准确度如何？误差的来源是什么？

How accurate is the estimate, and what is the source of the error?

估计误差仅为 9%。误差的产生是因为线性近似

The estimate is in error by only 9%. The error arises because the linear approximation

不包括分数变化 (Δx)/x 的平方（或更高次幂）（问题 5.17要求您找到平方项）。

does not include the square (or higher powers) of the fractional change (Δx)/x (Problem 5.17 asks you to find the squared term).

如何减少线性近似中的误差？

How can the error in the linear approximation be reduced?

为了减少误差，可以减小小数部分的变化。由于小数部分的变化取决于大数部分，因此我们来提高大数部分的精度。因此，将 1/13 乘以 8/8（1 的便捷形式），得到 8/104。其大数部分 0.08 已经可以近似于 1/13，误差在 4% 以内。为了改进精度，将 1/104 写为 (x + Δx) ⁻¹，其中 x = 100，Δx = 4。小数部分的变化 (Δx)/x 现在为 0.04（而不是 0.3）；对 1/x 和 8/x 的小数部分修正仅为 −4%。修正后的估计值为 0.0768：

To reduce the error, reduce the fractional change. Because the fractional change is determined by the big part, let’s increase the accuracy of the big part. Accordingly, multiply 1/13 by 8/8, a convenient form of 1, to construct 8/104. Its big part 0.08 approximates 1/13 already to within 4%. To improve it, write 1/104 as (x + Δx)⁻¹ with x = 100 and Δx = 4. The fractional change (Δx)/x is now 0.04 (rather than 0.3); and the fractional correction to 1/x and 8/x is a mere −4%. The corrected estimate is 0.0768:

这个估算可以在几秒钟内心算完成，并且精确到 0.13%！

This estimate can be done mentally in seconds and is accurate to 0.13%!

5.3.2 平方根

5.3.2 Square roots

分数指数 n = 1/2 提供了估算平方根的方法。例如，我们将其重写为 (x + Δx) ^1/2，其中 x = 9，Δx = 1。大数部分 x ^1/2为 3。由于 (Δx)/x = 1/9 且 n = 1/2，因此分数修正值为 1/18。修正后的估计值为

The fractional exponent n = 1/2 provides the method for estimating square roots. As an example, let’s estimate Rewrite it as (x + Δx)^1/2 with x = 9 and Δx = 1. The big part x^1/2 is 3. Because (Δx)/x = 1/9 and n = 1/2, the fractional correction is 1/18. The corrected estimate is

精确值为 3.1622...，因此估计精确到 0.14%。

The exact value is 3.1622 . . ., so the estimate is accurate to 0.14%.

5.3.3 季节变化的原因是什么？

5.3.3 A reason for the seasons?

人们常说，夏季比冬季温暖，是因为夏季地球比冬季更靠近太阳。这种常见的解释有两个原因，是站不住脚的。首先，尽管南半球与太阳的距离几乎没有差别，但南半球的夏季与北半球的冬季并存。其次，正如我们现在将要估计的那样，变化的地日距离产生的温差太小。距离决定太阳辐射强度，而辐射强度决定地表温度的因果关系，最容易用分形变化来分析。

Summers are warmer than winters, it is often alleged, because the earth is closer to the sun in the summer than in the winter. This common explanation is bogus for two reasons. First, summers in the southern hemisphere happen alongside winters in the northern hemisphere, despite almost no difference in the respective distances to the sun. Second, as we will now estimate, the varying earth–sun distance produces too small a temperature difference. The causal chain—that the distance determines the intensity of solar radiation and that the intensity determines the surface temperature—is most easily analyzed using fractional changes.

太阳辐射强度：强度等于太阳能量除以其辐射面积。太阳能量在一年内几乎不会变化（太阳已经存在了数十亿年）；然而，在距离太阳 r 处，能量已经扩散到一个表面积约为 r ²的巨大球体上。因此，强度 I 随 I α r ⁻²变化。半径和强度的分数变化关系如下：

Intensity of solar radiation: The intensity is the solar power divided by the area over which it spreads. The solar power hardly changes over a year (the sun has existed for several billion years); however, at a distance r from the sun, the energy has spread over a giant sphere with surface area ∼ r². The intensity I therefore varies according to I α r⁻². The fractional changes in radius and intensity are related by

地表温度：入射的太阳能无法累积，最终以黑体辐射的形式返回太空。根据斯特藩-玻尔兹曼定律 I = σT ⁴（习题 1.12），其输出强度取决于地球表面温度 T，其中 σ 为斯特藩-玻尔兹曼常数。因此，T α I ^1/4。使用分数变化，

Surface temperature: The incoming solar energy cannot accumulate and returns to space as blackbody radiation. Its outgoing intensity depends on the earth’s surface temperature T according to the Stefan–Boltzmann law I = σT⁴ (Problem 1.12), where σ is the Stefan–Boltzmann constant. Therefore T α I^1/4. Using fractional changes,

该关系式将强度与温度联系起来。温度与距离的关系式为 (ΔI)/I = −2 × (Δr)/r。将这两个关系式结合起来，距离与温度的关系如下：

This relation connects intensity and temperature. The temperature and distance are connected by (ΔI)/I = −2 × (Δr)/r. When joined, the two relations connect distance and temperature as follows:

计算的下一步是估算输入值 (Δr)/r，即地日距离的分数变化。地球绕太阳公转的轨道为椭圆形；其轨道距离为

The next step in the computation is to estimate the input (Δr)/r—namely, the fractional change in the earth–sun distance. The earth orbits the sun in an ellipse; its orbital distance is

其中ε是轨道偏心率，θ是极角，l是半直角。因此，r从r _min = l/(1 + ε)（当θ = 0°时）变化到r _{max = l/(1 − ε)（当θ = 180°时）。从r min}到l的增加导致了大约ε的微小变化。从l到r _max的增加又导致了大约ε的微小变化。因此，r的变化量大约为2ε。对于地球轨道而言，ε = 0.016，因此地日距离变化了0.032或3.2%（导致强度变化6.4%）。

where ε is the eccentricity of the orbit, θ is the polar angle, and l is the semilatus rectum. Thus r varies from r_min = l/(1 + ε) (when θ = 0°) to r_max = l/(1 − ε) (when θ = 180°). The increase from r_min to l contributes a fractional change of roughly ε. The increase from l to r_max contributes another fractional change of roughly ε. Thus, r varies by roughly 2ε. For the earth’s orbit, ε = 0.016, so the earth–sun distance varies by 0.032 or 3.2% (making the intensity vary by 6.4%).

距离增加 3.2% 会导致温度略微下降：

A 3.2% increase in distance causes a slight drop in temperature:

然而，人类并不仅仅依靠分数变化而生存，还经历绝对温度变化T。

However, man does not live by fractional changes alone and experiences the absolute temperature change T.

冬季T ≈ 0° C ，那么ΔT ≈ 0° C吗？

In winter T ≈ 0° C, so is ΔT ≈ 0° C?

如果我们的计算预测ΔT≈0 ° C，那它肯定是错的。如果用华氏度来测量T，就会得出更不合理的结论，因为在北半球的部分地区，T通常为负值。然而，ΔT不能仅仅因为T是以华氏度来测量的就改变符号！

If our calculation predicts that ΔT ≈ 0° C, it must be wrong. An even less plausible conclusion results from measuring T in Fahrenheit degrees, which makes T often negative in parts of the northern hemisphere. Yet ΔT cannot flip its sign just because T is measured in Fahrenheit degrees!

^{幸运的是，温标受斯特藩-玻尔兹曼定律约束。为了使黑体通量与T 4}成正比，温度必须相对于热能为零的状态（即绝对零度）进行测量。摄氏温标和华氏温标均不满足这一要求。

Fortunately, the temperature scale is constrained by the Stefan–Boltzmann law. For blackbody flux to be proportional to T⁴, temperature must be measured relative to a state with zero thermal energy: absolute zero. Neither the Celsius nor the Fahrenheit scale satisfies this requirement.

相比之下，开尔文温标测量的是相对于绝对零度的温度。在开尔文温标中，平均地表温度 T ≈ 300 K；因此，T 变化 1.6% 会导致 ΔT ≈ 5 K。5 K 的变化也意味着 5°C 的变化——开尔文和摄氏度大小相同，尽管这两个温标的零点不同。（另见问题 5.26。）温带地区夏季和冬季之间的典型温差为 20°C——即使考虑到估算中的误差，也远大于预测的 5°C 变化。日地距离变化对于季节变化的原因是一个令人怀疑的解释。

In contrast, the Kelvin scale does measure temperature relative to absolute zero. On the Kelvin scale, the average surface temperature is T ≈ 300 K; thus, a 1.6% change in T makes ΔT ≈ 5 K. A 5 K change is also a 5° C change—Kelvin and Celsius degrees are the same size, although the scales have different zero points. (See also Problem 5.26.) A typical temperature change between summer and winter in temperate latitudes is 20° C—much larger than the predicted 5° C change, even after allowing for errors in the estimate. A varying earth–sun distance is a dubious explanation of the reason for the seasons.

5.3.4 有效期限制

5.3.4 Limits of validity

线性分数变化近似

The linear fractional-change approximation

一直很有用。但它何时有效？为了避免被符号淹没，我们可以将 Δx 写为 z；然后取 x = 1，使 z 为绝对值和分数变化。右边变成 nz，线性分数变化近似等价于

has been useful. But when is it valid? To investigate without drowning in notation, write z for Δx; then choose x = 1 to make z the absolute and the fractional change. The right side becomes nz, and the linear fractional-change approximation is equivalent to

当 z 过大时，近似值会变得不准确：例如，当z = 1 时（问题 5.22）。指数 n 是否也受到限制？前面的例子仅展示了中等大小的指数：n = 2 表示能耗（第 5.2.3 节），-2 表示燃油效率（问题 5.18），-1 表示倒数（第 5.3.1 节），1/2 表示平方根（第 5.3.2 节），-2 和 1/4 表示季节（第 5.3.3 节）。我们需要更多数据。

The approximation becomes inaccurate when z is too large: for example, when evaluating with z = 1 (Problem 5.22). Is the exponent n also restricted? The preceding examples illustrated only moderate-sized exponents: n = 2 for energy consumption (Section 5.2.3), −2 for fuel efficiency (Problem 5.18), −1 for reciprocals (Section 5.3.1), 1/2 for square roots (Section 5.3.2), and −2 and 1/4 for the seasons (Section 5.3.3). We need further data.

在指数较大的极端情况下会发生什么？

What happens in the extreme case of large exponents?

对于较大的指数，例如 n = 100 和 z = 0.001，近似值预测 1.001 ¹⁰⁰ ≈ 1.1——接近真实值 1.105……然而，选择相同的 n 和 z = 0.1（大于 0.001 但仍然很小）会产生糟糕的预测

With a large exponent such as n = 100 and, say, z = 0.001, the approximation predicts that 1.001¹⁰⁰ ≈ 1.1—close to the true value of 1.105 . . . However, choosing the same n alongside z = 0.1 (larger than 0.001 but still small) produces the terrible prediction

1.1 ¹⁰⁰大约是 14,000，比预测值大 1000 多倍。

1.1¹⁰⁰ is roughly 14,000, more than 1000 times larger than the prediction.

两个预测都使用了较大的 n 和较小的 z，但只有一个预测准确；因此，问题不可能仅仅出在 n 或 z 上。罪魁祸首或许是无量纲乘积 nz。为了验证这一想法，请保持 nz 不变，同时尝试较大的 n 值。对于 nz 来说，合理的常数是 1——最简单的无量纲数。以下是几个例子。

Both predictions used large n and small z, yet only one prediction was accurate; thus, the problem cannot lie in n or z alone. Perhaps the culprit is the dimensionless product nz. To test that idea, hold nz constant while trying large values of n. For nz, a sensible constant is 1—the simplest dimensionless number. Here are several examples.

在每个例子中，近似值都错误地预测 (1 + z) ⁿ = 2。

In each example, the approximation incorrectly predicts that (1 + z)ⁿ = 2.

错误原因是什么？

What is the cause of the error?

为了找到原因，将序列延伸至 1.001 ¹⁰⁰⁰以上，并希望出现一种模式：这些值似乎接近于 e = 2.718281828……（自然对数的底数）。因此，对整个近似值取对数。

To find the cause, continue the sequence beyond 1.001¹⁰⁰⁰ and hope that a pattern will emerge: The values seem to approach e = 2.718281828 . . ., the base of the natural logarithms. Therefore, take the logarithm of the whole approximation.

图形推理表明，当 z 1 时，ln(1 + z) ≈ z （第 4.3 节）。因此，n ln(1 + z) ≈ nz，使得 (1 + z) ⁿ ≈ e ^nz。这种改进的近似值解释了为什么近似值 (1 + z) ⁿ ≈ 1 + nz 在 nz 较大时失效：只有当 nz 1 大约等于 e ^nz 1 + nz 时才成立。因此，当 z 1 时，两个最简单的近似值是

Pictorial reasoning showed that ln(1 + z) ≈ z when z 1 (Section 4.3). Thus, n ln(1 + z) ≈ nz, making (1 + z)ⁿ ≈ e^nz. This improved approximation explains why the approximation (1 + z)ⁿ ≈ 1 + nz failed with large nz: Only when nz 1 is e^nz approximately 1 + nz. Therefore, when z 1 the two simplest approximation are

该图显示了整个 n-z 平面上每个区域的最简近似。坐标轴为对数坐标，n 和 z 假设为正：右半平面 z 为1，上半平面 n 为1。右下角的边界曲线为 n ln z = 1。解释边界并扩展近似值是一项有益的练习（习题 5.28）。

The diagram shows, across the whole n–z plane, the simplest approximation in each region. The axes are logarithmic and n and z are assumed positive: The right half plane shows z 1, and the upper half plane shows n 1. On the lower right, the boundary curve is n ln z = 1. Explaining the boundaries and extending the approximations is an instructive exercise (Problem 5.28).

5.4 逐次逼近：井有多深？

5.4 Successive approximation: How deep is the well?

下一个取出大部分的说明强调了逐次逼近，并伪装成物理问题。

The next illustration of taking out the big part emphasizes successive approximation and is disguised as a physics problem.

你把一块石头扔进一口深度未知的井里，4秒后听到了水花飞溅的声音。忽略空气阻力，求出h，误差在5%以内。设c _s = 340 ms ⁻¹为声速，g = 10 ms ⁻²为重力强度。

You drop a stone down a well of unknown depth h and hear the splash 4 s later. Neglecting air resistance, find h to within 5%. Use c_s = 340 ms⁻¹ as the speed of sound and g = 10 ms⁻² as the strength of gravity.

近似解和精确解给出了几乎相同的井深，但却提供了截然不同的理解。

Approximate and exact solutions give almost the same well depth, but offer significantly different understandings.

5.4.1 精确深度

5.4.1 Exact depth

深度由以下约束决定：4 秒的等待时间分为两个时间：岩石自由落下井的时间和声音沿井向上传播的时间。自由落体时间为（问题 1.3），因此总时间为

The depth is determined by the constraint that the 4 s wait splits into two times: the rock falling freely down the well and the sound traveling up the well. The free-fall time is (Problem 1.3), so the total time is

为了精确求解 h，要么将平方根放在一边，然后对两边求平方，得到 h 的二次方程（问题 5.30）；或者，为了减少错误，将约束重写为新变量 z = 的二次方程。

To solve for h exactly, either isolate the square root on one side and square both sides to get a quadratic equation in h (Problem 5.30); or, for a less error-prone method, rewrite the constraint as a quadratic equation in a new variable z = .

作为 z = 的二次方程，约束为

As a quadratic equation in z = , the constraint is

使用二次公式并选择正根可得出

Using the quadratic formula and choosing the positive root yields

因为 z ² = h，

Because z² = h,

代入 g = 10 ms ⁻²和 c _s = 340 ms ⁻¹得出 h ≈ 71.56 m。

Substituting g = 10 ms⁻² and c_s = 340 ms⁻¹ gives h ≈ 71.56 m.

即使深度正确，其精确公式也一团糟。这种高熵的恐怖经常源于二次公式；它的使用往往标志着符号操控战胜了思维。我们会发现，精确答案可能不如近似答案有用。

Even if the depth is correct, the exact formula for it is a mess. Such high-entropy horrors arise frequently from the quadratic formula; its use often signals the triumph of symbol manipulation over thought. Exact answers, we will find, may be less useful than approximate answers.

5.4.2 大致深度

5.4.2 Approximate depth

要找到低熵的近似深度，需要确定最重要的部分——最重要的效应。这里，大部分时间是岩石的自由落体：即使岩石自由落体持续了4秒，其最大速度也只有gΤ = 40 ms ⁻¹，远低于c _s。因此，最重要的效应应该出现在声速无限大的极端情况下。

To find a low-entropy, approximate depth, identify the big part—the most important effect. Here, most of the total time is the rock’s free fall: The rock’s maximum speed, even if it fell for the entire 4 s, is only gΤ = 40 ms⁻¹, which is far below c_s. Therefore, the most important effect should arise in the extreme case of infinite sound speed.

如果c _s = ∞，那么井有多深？

If c_s = ∞, how deep is the well?

在这个零阶近似中，自由落体时间 t ₀为全时间 T = 4 s，因此井深 h ₀变为

In this zeroth approximation, the free-fall time t₀ is the full time T = 4 s, so the well depth h₀ becomes

这个近似深度是高估了还是低估了？它的准确度如何？

Is this approximate depth an overestimate or underestimate? How accurate is it?

这种近似值忽略了声音的传播时间，因此高估了自由落体的时间，从而也高估了深度。与大约71.56米的真实深度相比，它仅高估了11%——对于一种能够提供物理洞察力的快速方法来说，这个精度还算合理。此外，这种近似值本身也体现了其改进之处。

This approximation neglects the sound-travel time, so it overestimates the free-fall time and therefore the depth. Compared to the true depth of roughly 71.56 m, it overestimates the depth by only 11%—reasonable accuracy for a quick method offering physical insight. Furthermore, this approximation suggests its own refinement.

如何改进这种近似？

How can this approximation be improved?

为了改进它，使用近似深度 h ₀来近似声音传播时间。

To improve it, use the approximate depth h₀ to approximate the sound-travel time.

剩余时间是自由落体时间的下一个近似值。

The remaining time is the next approximation to the free-fall time.

在这段时间内，岩石下落了一段距离，因此深度的下一个近似值是

In that time, the rock falls a distance so the next approximation to the depth is

这个近似深度是高估了还是低估了？它的准确度如何？

Is this approximate depth an overestimate or underestimate? How accurate is it?

_{h 1}的计算使用了 h ₀来估算声音的传播时间。由于 h ₀高估了深度，因此该程序会高估声音的传播时间，同时低估自由落体时间，两者之差相同。因此，h ₁低估了深度。事实上，h ₁略小于大约 71.56 米的真实深度，但仅小了 1.3%。

The calculation of h₁ used h₀ to estimate the sound-travel time. Because h₀ overestimates the depth, the procedure overestimates the sound-travel time and, by the same amount, underestimates the free-fall time. Thus h₁ underestimates the depth. Indeed, h₁ is slightly smaller than the true depth of roughly 71.56 m—but by only 1.3%.

与精确求解二次公式相比，逐次逼近法有几个优点。首先，它有助于我们加深对系统的物理理解；例如，我们意识到T = 4秒的大部分时间都花在了自由落体上，因此深度大约为gT² ^/ 2。其次，它有图形化的解释（问题5.34）。第三，它能快速给出足够准确的答案。如果你想知道跳进井里是否安全，为什么要把深度计算到小数点后三位？

The method of successive approximation has several advantages over solving the quadratic formula exactly. First, it helps us develop a physical understanding of the system; we realize, for example, that most of the T = 4 s is spent in free fall, so the depth is roughly gT²/2. Second, it has a pictorial explanation (Problem 5.34). Third, it gives a suffciently accurate answer quickly. If you want to know whether it is safe to jump into the well, why calculate the depth to three decimal places?

最后，该方法可以处理模型中的微小变化。例如，声速可能随深度变化，或者空气阻力变得很重要（问题 5.32）。这时，强力的二次公式方法就失效了。二次公式，甚至更复杂的三次和四次公式，都是复杂方程的罕见闭式解。大多数方程没有闭式解。因此，如果我们要求一个精确的答案，对可解模型进行微小的改动通常就会产生一个难以处理的模型。逐次逼近法是一种稳健的替代方法，可以产生低熵、易于理解的解。

Finally, the method can handle small changes in the model. Maybe the speed of sound varies with depth, or air resistance becomes important (Problem 5.32). Then the brute-force, quadratic-formula method fails. The quadratic formula and the even messier cubic and the quartic formulas are rare closed-form solutions to complicated equations. Most equations have no closed-form solution. Therefore, a small change to a solvable model usually produces an intractable model—if we demand an exact answer. The method of successive approximation is a robust alternative that produces low-entropy, comprehensible solutions.

5.5 令人畏惧的三角积分

5.5 Daunting trigonometric integral

最后一个例子是估算一个令人望而生畏的三角积分，这是我本科时学过的。我和同学们经常在物理图书馆熬夜解答作业；研究生们也一样，用他们最喜欢的数学和物理问题来招待我们。

The final example of taking out the big part is to estimate a daunting trigonometric integral that I learned as an undergraduate. My classmates and I spent many late nights in the physics library solving homework problems; the graduate students, doing the same for their courses, would regale us with their favorite mathematics and physics problems.

这个积分出现在前苏联朗道理论物理研究所的数学预科考试中。题目要求计算

The integral appeared on the mathematical-preliminaries exam to enter the Landau Institute for Theoretical Physics in the former USSR. The problem is to evaluate

无需使用计算器或计算机，在不到 5 分钟的时间内即可达到 5% 以内！

to within 5% in less than 5 min without using a calculator or computer!

(cos t) ¹⁰⁰看起来吓人。大多数三角恒等式都帮不上忙。通常有用的恒等式 (cos t) ² = (cos 2t − 1)/2 只会产生

That (cos t)¹⁰⁰ looks frightening. Most trigonometric identities do not help. The usually helpful identity (cos t)² = (cos 2t − 1)/2 produces only

它在展开 50 次方后就变成了一个三角函数怪物。

which becomes a trigonometric monster upon expanding the 50th power.

一个指向更简单方法的线索是，5% 的精度就足够了——所以，找到最大的部分！当 t 接近于零时，被积函数最大。此时，成本 t ≈ 1 − t ² /2 （问题 5.20），因此被积函数大致为

A clue pointing to a simpler method is that 5% accuracy is sufficient—so, find the big part! The integrand is largest when t is near zero. There, cos t ≈ 1 − t²/2 (Problem 5.20), so the integrand is roughly

它具有我们熟悉的形式 (1 + z) ⁿ，其中小数变化 z = −t ² /2 ，指数 n = 100。当 t 很小时，z = −t ² /2 很小，因此 (1 + z) ^{n可以使用}第 5.3.4 节的结果来近似：

It has the familiar form (1 + z)ⁿ, with fractional change z = −t²/2 and exponent n = 100. When t is small, z = −t²/2 is tiny, so (1 + z)ⁿ may be approximated using the results of Section 5.3.4:

由于指数 n 很大，即使 t 和 z 较小，nz 也可能很大。因此，最安全的近似值为 (1 + z) ⁿ ≈ e ^nz；则

Because the exponent n is large, nz can be large even when t and z are small. Therefore, the safest approximation is (1 + z)ⁿ ≈ e^nz; then

余弦的高次幂变为高斯！

A cosine raised to a high power becomes a Gaussian!

为了检验这一令人惊讶的结论，计算机生成的 (cost t) ⁿ (n = 1 ... 5) 图显示，随着 n 的增加，形成了高斯钟形。

As a check on this surprising conclusion, computer-generated plots of (cos t)ⁿ for n = 1 . . . 5 show a Gaussian bell shape taking form as n increases.

即使有这样的图形证据，用高斯函数代替 (cost t) ¹⁰⁰也有点可疑。在原始积分中，t 的范围从 −π/2 到 π/2，而这些端点远远超出了 cost t ≈ 1 − t ² /2 的精确近似值区域。幸运的是，这个问题只会导致一个很小的误差（问题 5.35）。忽略这个误差会将原始积分变成具有有限极限的高斯积分：

Even with this graphical evidence, replacing (cos t)¹⁰⁰ by a Gaussian is a bit suspicious. In the original integral, t ranges from −π/2 to π/2, and these endpoints are far outside the region where cos t ≈ 1 − t²/2 is an accurate approximation. Fortunately, this issue contributes only a tiny error (Problem 5.35). Ignoring this error turns the original integral into a Gaussian integral with finite limits:

不幸的是，在有限极限下，积分没有闭合形式。但将极限扩展到无穷大可以得到一个闭合形式，而且几乎不产生误差（问题 5.36）。近似链现在是

Unfortunately, with finite limits the integral has no closed form. But extending the limits to infinity produces a closed form while contributing almost no error (Problem 5.36). The approximation chain is now

最后一个积分是老朋友了（第 2.1 节）：= 。当 α = 50 时，积分变为。方便的是，50 大约等于 16π，因此平方根（也就是我们 5% 的估计值）大约等于 0.25。

The last integral is an old friend (Section 2.1): = . With α = 50, the integral becomes . Conveniently, 50 is roughly 16π, so the square root—and our 5% estimate—is roughly 0.25.

为了比较，精确积分是（问题 5.41）

For comparison, the exact integral is (Problem 5.41)

当 n = 100 时，二项式系数和二的幂产生

When n = 100, the binomial coeffcient and power of two produce

我们对 5 分钟内 5% 以内的估计为 0.25，精确到几乎 0.01%！

Our 5-minute, within-5% estimate of 0.25 is accurate to almost 0.01%!

5.6 总结和进一步的问题

5.6 Summary and further problems

遇到复杂问题时，将其分解成一个主要部分（最重要的影响）和一个修正部分。先分析主要部分，然后再考虑修正部分。这种逐次逼近法，一种分而治之的推理，会自动以低熵形式给出结果。低熵表达式几乎没有合理的替代方案；因此，它们更容易记忆和理解。简而言之，近似结果可能比精确结果更有用。

Upon meeting a complicated problem, divide it into a big part—the most important effect—and a correction. Analyze the big part first, and worry about the correction afterward. This successive-approximation approach, a species of divide-and-conquer reasoning, gives results automatically in a low-entropy form. Low-entropy expressions admit few plausible alternatives; they are therefore memorable and comprehensible. In short, approximate results can be more useful than exact results.

6

6

类比

Analogy

处境艰难时，强者会降低标准。这一理念，也就是整本书的主题，是最后一个街头格斗工具——类比推理的基础。它的建议很简单：面对一个难题，构造并解决一个类似但更简单的问题——一个类比问题。熟能生巧。该工具在空间三角学（第 6.1 节）中引入；在立体几何和拓扑学（第 6.2 节）中得到进一步完善；然后应用于离散数学（第 6.3 节），并在最后的例子中应用于无穷超越和（第 6.4 节）。

When the going gets tough, the tough lower their standards. This idea, the theme of the whole book, underlies the final street-fighting tool of reasoning by analogy. Its advice is simple: Faced with a difficult problem, construct and solve a similar but simpler problem—an analogous problem. Practice develops fluency. The tool is introduced in spatial trigonometry (Section 6.1); sharpened on solid geometry and topology (Section 6.2); then applied to discrete mathematics (Section 6.3) and, in the farewell example, to an infinite transcendental sum (Section 6.4).

6.1 空间三角学：甲烷中的键角

6.1 Spatial trigonometry: The bond angle in methane

第一个类比来自空间三角学。在甲烷（化学式 CH4 _）中，一个碳原子位于正四面体的中心，每个顶点各有一个氢原子。两个碳氢键之间的夹角 θ 是多少？

The first analogy comes from spatial trigonometry. In methane (chemical formula CH₄), a carbon atom sits at the center of a regular tetrahedron, and one hydrogen atom sits at each vertex. What is the angle θ between two carbon–hydrogen bonds?

三维空间中的角度很难直观地表示。例如，试着想象并计算正四面体两个面之间的角度。由于二维角度很容易直观地表示，让我们构建并分析一个类似的平面分子。了解它的键角或许有助于我们猜测甲烷的键角。

Angles in three dimensions are hard to visualize. Try, for example, to imagine and calculate the angle between two faces of a regular tetrahedron. Because two-dimensional angles are easy to visualize, let’s construct and analyze an analogous planar molecule. Knowing its bond angle might help us guess methane’s bond angle.

类似的平面分子应该有四个还是三个氢原子？

Should the analogous planar molecule have four or three hydrogens?

四个氢原子形成四个键，当它们在平面上规则排列时，会产生两个不同的键角。相比之下，甲烷只有一个键角。因此，使用四个氢原子会改变原始问题的一个关键特征。可能的解决办法是构建类似的仅使用三个氢原子的平面分子。

Four hydrogens produce four bonds which, when spaced regularly in a plane, produce two different bond angles. In contrast, methane contains only one bond angle. Therefore, using four hydrogens alters a crucial feature of the original problem. The likely solution is to construct the analogous planar molecule using only three hydrogens.

三个氢原子规则地排列在一个平面上，只会产生一个键角：θ = 120°。或许这个键角就是甲烷的键角！然而，一个数据点对于更高维度的预测来说，只是纸上谈兵。二维空间（d = 2）的单个数据点与许多猜想相符——例如，在d维空间中，键角是120°、(60d)°，或者许多其他值。

Three hydrogens arranged regularly in a plane create only one bond angle: θ = 120°. Perhaps this angle is the bond angle in methane! One data point, however, is a thin reed on which to hang a prediction for higher dimensions. The single data point for two dimensions (d = 2) is consistent with numerous conjectures—for example, that in d dimensions the bond angle is 120° or (60d)° or much else.

选择一个合理的猜想需要收集更多数据。一个更简单但更类似的问题很容易获得数据：一维线性分子 CH2 _。它的两个氢原子彼此相对，因此两个 C–H 键形成 θ = 180° 的夹角。

Selecting a reasonable conjecture requires gathering further data. Easily available data comes from an even simpler yet analogous problem: the one-dimensional, linear molecule CH₂. Its two hydrogens sit opposite one another, so the two C–H bonds form an angle of θ = 180°.

根据积累的数据，对于三维角度θ ₃有哪些合理的推测？

Based on the accumulated data, what are reasonable conjectures for the three-dimensional angle θ₃?

一维分子排除了θ _d = (60d)°的猜想。它还提出了新的猜想，例如θ _d = (240 − 60d)°或θ _d = 360°/(d + 1)。检验这些猜想是简单情况方法的理想任务。高维（高维）简单情况检验推翻了θ _d = (240 − 60d)°的猜想。对于高维分子，它预测了不合理的键角，即当d = 4时θ = 0，当d > 4时θ < 0。

The one-dimensional molecule eliminates the conjecture that θ_d = (60d)°. It also suggests new conjectures—for example, that θ_d = (240 − 60d)° or θ_d = 360°/(d + 1). Testing these conjectures is an ideal task for the method of easy cases. The easy-cases test of higher dimensions (high d) refutes the conjecture that θ_d = (240 − 60d)°. For high d, it predicts implausible bond angles—namely, θ = 0 for d = 4 and θ < 0 for d > 4.

幸运的是，第二个假设 θ _d = 360°/(d + 1) 通过了同样的简单情况检验。让我们继续验证它对甲烷的预测——即 θ ₃ = 90°。想象一下甲烷的一个“大哥”：一个 CH ₆分子，碳原子位于立方体的中心，六个氢原子位于立方体的中心。它的小键角为 90°。（另一个键角为 180°。）现在移除两个氢原子，将 CH ₆变成 CH _{4 ，使剩下的四个氢原子均匀分布。减少拥挤度会使小键角超过 90°，从而推翻 θ 3} = 90°的预测。

Fortunately, the second suggestion, θ_d = 360°/(d + 1), passes the same easy-cases test. Let’s continue to test it by evaluating its prediction for methane—namely, θ₃ = 90°. Imagine then a big brother of methane: a CH₆ molecule with carbon at the center of a cube and six hydrogens at the face centers. Its small bond angle is 90°. (The other bond angle is 180°.) Now remove two hydrogens to turn CH₆ into CH₄, evenly spreading out the remaining four hydrogens. Reducing the crowding raises the small bond angle above 90°—and refutes the prediction that θ₃ = 90°.

迄今为止的数据已经推翻了最简单的有理函数猜想 (240−60d)° 和 360°/(d+1)。虽然其他有理函数猜想可能仍然存在，但只有两个数据点，可能性实在太大。更糟糕的是，θ _d甚至可能不是 d 的有理函数。

The data so far have refuted the simplest rational-function conjectures (240−60d)° and 360°/(d+1). Although other rational-function conjectures might survive, with only two data points the possibilities are too vast. Worse, θ_d might not even be a rational function of d.

进步需要新的思路：键角或许并非最容易研究的变量。在推测3、5、11、29……系列中的下一个项时，也会出现类似的困难。

Progress requires a new idea: The bond angle might not be the simplest variable to study. An analogous difficulty arises when conjecturing the next term in the series 3, 5, 11, 29, . . .

该系列的下一个术语是什么？

What is the next term in the series?

乍一看，这些数字似乎几乎是随机的。然而，从每一项中减去2，得到的结果是1、3、9、27……。因此，原始序列的下一项很可能是83。同样，对θ _{d数据进行简单的变换，或许能帮助我们推测θ d}的规律。

At first glance, the numbers seems almost random. Yet subtracting 2 from each term produces 1, 3, 9, 27, . . . Thus, in the original series the next term is likely to be 83. Similarly, a simple transformation of the θ_d data might help us conjecture a pattern for θ_d.

θ _d数据 的哪种变换会产生简单的模式？

What transformation of the θ_d data produces simple patterns?

所需的变换应该产生简单的图案，并具有美学或逻辑上的合理性。一个合理性是诚实计算键角的结构，它可以作为两个 C–H 向量的点积来计算（问题 6.3）。由于点积涉及余弦，因此 θ _d的一个合理变换是 cos θ _d。

The desired transformation should produce simple patterns and have aesthetic or logical justification. One justification is the structure of an honest calculation of the bond angle, which can be computed as a dot product of two C–H vectors (Problem 6.3). Because dot products involve cosines, a worthwhile transformation of θ_d is cos θ_d.

这种变换简化了数据：cos θ _d系列简单地以 −1、−1/2、... 开始。两个可能的延续是 −1/4 或 −1/3；它们分别对应于一般项 −1/2 ^d−1或 −1/d。

This transformation simplifies the data: The cos θ_d series begins simply −1, −1/2, . . . Two plausible continuations are −1/4 or −1/3; they correspond, respectively, to the general term −1/2^d−1 or −1/d.

哪种延续和猜想更合理？

Which continuation and conjecture is the more plausible?

两个猜想都预测 cosθ < 0，因此 θd _> 90°（对于所有 d）。这个共同的预测令人鼓舞（问题 6.1）；然而，共同的预测意味着它无法区分这两个猜想。

Both conjectures predict cos θ < 0 and therefore θ_d > 90° (for all d). This shared prediction is encouraging (Problem 6.1); however, being shared means that it does not distinguish between the conjectures.

这两种猜想是否符合分子几何构型？除了键角之外，一个重要的几何特征是碳的位置。在一个维度上，它位于中间两个氢原子之间，因此它将 H–H 线段分成长度比为 1:1 的两段。

Does either conjecture match the molecular geometry? An important geometric feature, apart from the bond angle, is the position of the carbon. In one dimension, it lies halfway between the two hydrogens, so it splits the H–H line segment into two pieces having a 1:1 length ratio.

在二维空间中，碳原子位于连接一个氢原子和另外两个氢原子中点的高度上，碳原子将该高度分成两部分，长度比为 1:2。

In two dimensions, the carbon lies on the altitude that connects one hydrogen to the midpoint of the other two hydrogens. The carbon splits the altitude into two pieces having a 1:2 length ratio.

碳如何分解出与甲烷类似的高度？

How does the carbon split the analogous altitude of methane?

在甲烷中，类似的高度从顶部顶点延伸至底部中心。碳原子位于平均位置，因此也位于四个氢原子的平均高度。由于三个底部氢原子的高度为零，因此四个氢原子的平均高度为h/4，其中h是顶部氢原子的高度。因此，在三维空间中，碳原子将高度分成两部分，长度比为h/4:3h/4，即1:3。因此，在d维空间中，碳原子可能将高度分成两部分，长度比为1:d（问题6.2）。

In methane, the analogous altitude runs from the top vertex to the center of the base. The carbon lies at the mean position and therefore at the mean height of the four hydrogens. Because the three base hydrogens have zero height, the mean height of the four hydrogens is h/4, where h is the height of the top hydrogen. Thus, in three dimensions, the carbon splits the altitude into two parts having a length ratio of h/4 : 3h/4 or 1 : 3. In d dimensions, therefore, the carbon probably splits the altitude into two parts having a length ratio of 1:d (Problem 6.2).

由于 1 : d 在几何中自然成立，cos θ _d更有可能包含 1/d 而不是 1/2 ^d−1。因此，两个 cos θ _d猜想中，可能性更大的是

Because 1 : d arises naturally in the geometry, cos θ_d is more likely to contain 1/d rather than 1/2^d−1. Thus, the more likely of the two cos θ_d conjectures is that

对于甲烷，当 d = 3 时，预测的键角为 arccos(−1/3)，约为 109.47°。这个基于类比推理的预测结果与实验结果以及基于解析几何的诚实计算结果（问题 6.3）相符。

For methane, where d = 3, the predicted bond angle is arccos(−1/3) or approximately 109.47°. This prediction using reasoning by analogy agrees with experiment and with an honest calculation using analytic geometry (Problem 6.3).

6.2 拓扑：有多少个区域？

6.2 Topology: How many regions?

甲烷中的键角（第 6.1 节）可以直接用解析几何计算（问题 6.3），因此类比推理并不能充分发挥其威力。因此，请尝试以下问题。

The bond angle in methane (Section 6.1) can be calculated directly with analytic geometry (Problem 6.3), so reasoning by analogy does not show its full power. Therefore, try the following problem.

五个平面将空间划分为多少个区域？

Into how many regions do five planes divide space?

此公式允许退化排列，例如五个平行平面、四个平面交于一点或三个平面交于一条线。为了消除这些及其他退化情况，我们随机放置和调整平面的方向，从而最大化区域数量。问题在于求出五个平面所能生成的区域的最大数量。

This formulation permits degenerate arrangements such as five parallel planes, four planes meeting at a point, or three planes meeting at a line. To eliminate these and other degeneracies, let’s place and orient the planes randomly, thereby maximizing the number of regions. The problem is then to find the maximum number of regions produced by five planes.

五个平面很难想象，但简单情况（使用更少的平面）的方法或许能得出一个推广到五个平面的模式。最简单的情况是零个平面：空间保持完整，因此 R(0) = 1（其中 R(n) 表示 n 个平面产生的区域数）。第一个平面将空间分成两半，得出 R(1) = 2。要添加第二个平面，想象一下将橙子切成两片，得到四个楔形：R(2) = 4。

Five planes are hard to imagine, but the method of easy cases—using fewer planes—might produce a pattern that generalizes to five planes. The easiest case is zero planes: Space remains whole so R(0) = 1 (where R(n) denotes the number of regions produced by n planes). The first plane divides space into two halves, giving R(1) = 2. To add the second plane, imagine slicing an orange twice to produce four wedges: R(2) = 4.

数据中出现了什么模式？

What pattern(s) appear in the data?

一个合理的推测是 R(n) = 2n ^。为了验证这一推测，我们可以尝试 n = 3 的情况，将橙子切三遍，将四块橙子分别切成两小块；这样，R(3) 确实等于 8。或许，同样的规律在 R(4) = 16 和 R(5) = 32 时仍然成立。在下表中，这两个外推结果被标记为灰色，以区别于已验证的数值。

A reasonable conjecture is that R(n) = 2ⁿ. To test it, try the case n = 3 by slicing the orange a third time and cutting each of the four pieces into two smaller pieces; thus, R(3) is indeed 8. Perhaps the pattern continues with R(4) = 16 and R(5) = 32. In the following table for R(n), these two extrapolations are marked in gray to distinguish them from the verified entries.

如何进一步检验R(n) = 2 ^n猜想？

How can the R(n) = 2ⁿ conjecture be tested further?

通过计数区域进行直接检验很困难，因为这些区域在三维空间中难以可视化。一个类似的二维问题更容易解决，其解法或许有助于检验三维猜想。二维空间由线分割，因此类似的问题如下：

A direct test by counting regions is difficult because the regions are hard to visualize in three dimensions. An analogous two-dimensional problem would be easier to solve, and its solution may help test the three-dimensional conjecture. A two-dimensional space is partitioned by lines, so the analogous question is the following:

n条线将平面最多划分成多少个区域？

What is the maximum number of regions into which n lines divide the plane?

简单案例法或许能揭示出一种模式。如果模式是 2 ⁿ，那么 R(n) = 2 ⁿ猜想很可能在三维空间中适用。

The method of easy cases might suggest a pattern. If the pattern is 2ⁿ, then the R(n) = 2ⁿ conjecture is likely to apply in three dimensions.

在一些简单的情况下会发生什么？

What happens in a few easy cases?

零线离开平面完整，给出 R(0) = 1。接下来的三种情况如下（尽管参见问题 6.5）：

Zero lines leave the plane whole, giving R(0) = 1. The next three cases are as follows (although see Problem 6.5):

在 R(3) 被证明为 7 之前，R(n) = 2 ⁿ猜想看起来是成立的。然而，在放弃这样一个简单的猜想之前，画一条第四条线，仔细数一数其中的区域。四条线只构成了 11 个区域，而不是预测的 16 个，因此 2 ⁿ猜想已经失效。

Until R(3) turned out to be 7, the conjecture R(n) = 2ⁿ looked sound. However, before discarding such a simple conjecture, draw a fourth line and carefully count the regions. Four lines make only 11 regions rather than the predicted 16, so the 2ⁿ conjecture is dead.

当看到二维数据 R ₂ (n) 和三维数据 R ₃ (n) 时，可能会产生新的猜想。

A new conjecture might arise from seeing the two-dimensional data R₂(n) alongside the three-dimensional data R₃(n).

在此表中，多个条目组合起来形成相近的条目。例如，R ₂ (1) 和 R ₃ (1)（n = 1 列中的两个条目）加起来等于 R ₂ (2) 或 R ₃ (2)。这两个条目又加起来等于 R ₃ (3)。但是，该表中有许多小数，并且有多种组合方式；要排除这些巧合，需要收集更多数据——最简单的数据源就是类似的一维问题。

In this table, several entries combine to make nearby entries. For example, R₂(1) and R₃(1)—the two entries in the n = 1 column—sum to R₂(2) or R₃(2). These two entries in turn sum to the R₃(3) entry. But the table has many small numbers with many ways to combine them; discarding the coincidences requires gathering further data—and the simplest data source is the analogous one-dimensional problem.

n 个点将一条线最多分成多少个线段？

What is the maximum number of segments into which n points divide a line?

一个诱人的答案是，n 个点组成 n 条线段。然而，一个简单的情况——一个点组成两条线段——就降低了这个答案的吸引力。更确切地说，n 个点组成 n + 1 条线段。这个结果生成了下表中的R _1行。

A tempting answer is that n points make n segments. However, an easy case—that one point produces two segments—reduces the temptation. Rather, n points make n + 1 segments. That result generates the R₁ row in the following table.

这些数据中存在什么模式？

What patterns are in these data?

2n猜想部分成立。在 R1_行中，它从 n = 2 开始失效。在 R2_行中，它从 n = 3 开始失效。因此，在 R3 行中^，_它很可能从 n = 4 开始失效，这使得 R3(4) = 16 和 R3(5) = 32 这两个猜想_不太可能_成立。我个人估计，在发现这些猜想失效之前，R3(4) = 16 猜想的概率_{为0.5；但现在它最多降至 0.01。（更多关于估计和更新猜想概率的信息，请参阅 Corfield [} 11 ]、Jaynes [ 21 ] 和 Polya [ 36 ]关于合情推理的重要著作。）

The 2ⁿ conjecture survives partially. In the R₁ row, it fails starting at n = 2. In the R₂ row, it fails starting at n = 3. Thus in the R₃ row, it probably fails starting at n = 4, making the conjectures R₃(4) = 16 and R₃(5) = 32 improbable. My personal estimate is that, before seeing these failures, the probability of the R₃(4) = 16 conjecture was 0.5; but now it falls to at most 0.01. (For more on estimating and updating the probabilities of conjectures, see the important works on plausible reasoning by Corfield [11], Jaynes [21], and Polya [36].)

更好的消息是，这些明显的巧合蕴含着一个稳健的模式：

In better news, the apparent coincidences contain a robust pattern:

如果按照这个模式继续下去，五个平面可以将空间划分成多少个区域？

If the pattern continues, into how many regions can five planes divide space?

根据模式，

According to the pattern,

进而

and then

因此，五个平面最多可以将空间划分为26个区域。

Thus, five planes can divide space into a maximum of 26 regions.

通过绘制五个平面并计算区域数来推导这个数字非常困难。此外，这种蛮力推导只能得出 R ₃ (5) 的值，而简单的情况和类比方法可以计算表中的任意项。因此，它们提供了足够的数据来推测 R ₂ (n)（问题 6.9）、R ₃ (n)（问题 6.10）以及一般项 R _d (n)（问题 6.12）的表达式。

This number is hard to deduce by drawing five planes and counting the regions. Furthermore, that brute-force approach would give the value of only R₃(5), whereas easy cases and analogy give a method to compute any entry in the table. They thereby provide enough data to conjecture expressions for R₂(n) (Problem 6.9), for R₃(n) (Problem 6.10), and for the general entry R_d(n) (Problem 6.12).

6.3 算子：欧拉-麦克劳林求和

6.3 Operators: Euler–MacLaurin summation

下一个类比研究的是一些不常见的函数。大多数函数会将数字转换为其他数字，但有些特殊类型的函数——运算符——会将函数转换为其他函数。一个常见的例子是导数运算符 D。它将正弦函数转换为余弦函数，或将双曲正弦函数转换为双曲余弦函数。在运算符符号中，D(sin) = cos 和 D(sin h) = cosh；省略括号可以得到更简洁的表达式 D sin = cos 和 D sin h = cosh。要理解和学习如何使用运算符，一个有效的工具是类比推理：运算符的行为与普通函数非常相似，甚至与数字也类似。

The next analogy studies unusual functions. Most functions turn numbers into other numbers, but special kinds of functions—operators—turn functions into other functions. A familiar example is the derivative operator D. It turns the sine function into the cosine function, or the hyperbolic sine function into the hyperbolic cosine function. In operator notation, D(sin) = cos and D(sin h) = cosh; omitting the parentheses gives the less cluttered expression D sin = cos and D sin h = cosh. To understand and learn how to use operators, a fruitful tool is reasoning by analogy: Operators behave much like ordinary functions or even like numbers.

6.3.1 左移

6.3.1 Left shift

与数字类似，导数算子 D 可以平方得到 D ²（二阶导数算子），也可以得到 D 的任意整数幂。同样，导数算子也可以用于多项式。在这种情况下，一个普通的多项式，例如 P(x) = x ² + x/10 + 1，可以得到算子多项式 P(D) = D ² + D/10 + 1（小阻尼弹簧质量系统的微分算子）。

Like a number, the derivative operator D can be squared to make D² (the second-derivative operator) or to make any integer power of D. Similarly, the derivative operator can be fed to a polynomial. In that usage, an ordinary polynomial such as P(x) = x² + x/10 + 1 produces the operator polynomial P(D) = D² + D/10 + 1 (the differential operator for a lightly damped spring–mass system).

与数字的类比能延伸到什么程度？例如，cosh D 或 sin D 有意义吗？因为这些函数可以用指数函数写成，所以我们来研究一下指数 e ^D运算符。

How far does the analogy to numbers extend? For example, do cosh D or sin D have a meaning? Because these functions can be written using the exponential function, let’s investigate the operator exponential e^D.

e ^D 是什么意思？

What does e^D mean?

^{e D}的直接解释是，它将函数 f 转换为 e ^Df。

The direct interpretation of e^D is that it turns a function f into e^Df.

然而，这种解释毫无必要地是非线性的。它将 2f 转化为 e ^{2Df ，即 e Df}的平方，而一个由 f 生成 e ^{Df 的}线性算子，会由 2f 生成 2e ^Df。为了得到线性解释，可以使用泰勒级数（假设 D 是一个数字）来从线性算子中构造 e ^{D 。}

However, this interpretation is needlessly nonlinear. It turns 2f into e^2Df, which is the square of e^Df, whereas a linear operator that produces e^Df from f would produce 2e^Df from 2f. To get a linear interpretation, use a Taylor series—as if D were a number—to build e^D out of linear operators.

这个e ^D 对简单函数有什么作用？

What does this e^D do to simple functions?

最简单的非零函数是常数函数 f = 1。这是将该函数输入到 e ^D中：

The simplest nonzero function is the constant function f = 1. Here is that function being fed to e^D:

下一个最简单的函数 x 变成 x + 1。

The next simplest function x turns into x + 1.

更有趣的是，x ²变成 (x + 1) ²。

More interestingly, x² turns into (x + 1)².

e ^D 通常做什么？

What does e^D do in general?

^{前面的例子遵循 e D} x ⁿ = (x+1) ⁿ的模式。由于 x 的大多数函数都可以展开为 x 的幂，并且 e ^D将 x ^{n 的}每一项都转换为 (x + 1) ⁿ，因此结论是 e ^D将 f(x) 转换为 f(x + 1)。令人惊讶的是，e ^D就是 L，即左移运算符。

The preceding examples follow the pattern e^Dxⁿ = (x+1)ⁿ. Because most functions of x can be expanded in powers of x, and e^D turns each xⁿ term into (x + 1)ⁿ, the conclusion is that e^D turns f(x) into f(x + 1). Amazingly, e^D is simply L, the left-shift operator.

6.3.2 总结

6.3.2 Summation

正如导数算子可以表示左移算子（如 L = e ^D），左移算子也可以表示求和运算。这种算子表示将带来一种强大的方法，用于近似没有封闭形式的和。

Just as the derivative operator can represent the left-shift operator (as L = e^D), the left-shift operator can represent the operation of summation. This operator representation will lead to a powerful method for approximating sums with no closed form.

求和类似于我们更熟悉的积分运算。积分有定积分和不定积分两种形式：定积分等价于不定积分，然后在积分极限处求值。例如，这是 f(x) = 2x 的定积分。

Summation is analogous to the more familiar operation of integration. Integration occurs in definite and indefinite flavors: Definite integration is equivalent to indefinite integration followed by evaluation at the limits of integration. As an example, here is the definite integration of f(x) = 2x.

一般来说，输入函数 g 与不定积分结果之间的关系是 DG = g，其中 D 是导数算子，G = ∫g 是不定积分的结果。因此，D 和 ∫ 互为逆函数彼此之间——D ∫ = 1 或 D = 1/∫——图中的环路表示这种联系。（由于可能存在积分常数，∫D ≠ 1。）

In general, the connection between an input function g and the result of indefinite integration is DG = g, where D is the derivative operator and G = ∫g is the result of indefinite integration. Thus D and ∫ are inverses of one another—D ∫ = 1 or D = 1/∫—a connection represented by the loop in the diagram. (∫D ≠ 1 because of a possible integration constant.)

总结的类似图像是什么？

What is the analogous picture for summation?

与积分类似，将定积分定义为不定积分，然后在极限处求值。但应用类比时要谨慎，以免出现偏差一或栅栏柱误差（问题 2.24）。该和包含三个矩形——f(2)、f(3) 和 f(4)，而定积分dk 不包含 f(4) 的任何矩形。与其通过重新定义我们熟悉的积分运算来纠正这种差异，不如将不定积分解释为排除最后一个矩形。然后，进行不定积分，并在极限 a 和 b 处求值，得到的和的指标范围从 a 到 b − 1。

Analogously to integration, define definite summation as indefinite summation and then evaluation at the limits. But apply the analogy with care to avoid an off-by-one or fencepost error (Problem 2.24). The sum includes three rectangles—f(2), f(3), and f(4)—whereas the definite integral dk does not include any of the f(4) rectangle. Rather than rectifying the discrepancy by redefining the familiar operation of integration, interpret indefinite summation to exclude the last rectangle. Then indefinite summation followed by evaluating at the limits a and b produces a sum whose index ranges from a to b − 1.

举个例子，取 f(k) = k。那么不定和 Σf 就是函数 F，其定义为 F(k) = k(k−1)/2+C（其中 C 是求和常数）。在 0 和 n 之间求 F 的值可得出 n(n − 1)/2，即。在下图中，这些步骤是正向路径。

As an example, take f(k) = k. Then the indefinite sum Σf is the function F defined by F(k) = k(k−1)/2+C (where C is the constant of summation). Evaluating F between 0 and n gives n(n − 1)/2, which is . In the following diagram, these steps are the forward path.

在逆向路径中，新的 Δ 算子使 Σ 反转，就像微分使积分反转一样。因此，Δ 的算子表示形式也为 Σ 提供了表示形式。由于 Δ 和导数算子 D 类似，因此它们的表示形式可能也类似。导数是极限

In the reverse path, the new Δ operator inverts Σ just as differentiation inverts integration. Therefore, an operator representation for Δ provides one for Σ. Because Δ and the derivative operator D are analogous, their representations are probably analogous. A derivative is the limit

因此，导数算子 D 是算子极限

The derivative operator D is therefore the operator limit

其中 L _h运算符将 f(x) 转换为 f(x + h)，即 L _h左移 h。

where the L_h operator turns f(x) into f(x + h)—that is, L_h left shifts by h.

Δ 的类似表示是什么？

What is an analogous representation of Δ?

D 的算子极限使用无穷小左移；相应地，积分的逆运算是对无穷小宽度的矩形求和。由于求和 Σ 对单位宽度的矩形求和，其逆 Δ 应该使用单位左移，即 L _h，其中 h = 1。合理的推测是：

The operator limit for D uses an infinitesimal left shift; correspondingly, the inverse operation of integration sums rectangles of infinitesimal width. Because summation Σ sums rectangles of unit width, its inverse Δ should use a unit left shift—namely, L_h with h = 1. As a reasonable conjecture,

这个 Δ（称为有限差分算子）被构造为 1/Σ。如果构造正确，则 (L − 1)Σ 是恒等算子 1。换句话说，(L − 1)Σ 应该将函数转化为自身。

This Δ—called the finite-difference operator—is constructed to be 1/Σ. If the construction is correct, then (L − 1)Σ is the identity operator 1. In other words, (L − 1)Σ should turn functions into themselves.

这个猜想在各种简单情况下效果如何？

How well does this conjecture work in various easy cases?

为了检验这个猜想，首先将算子 (L−1)Σ 应用于简单函数 g = 1。然后 Σg 是一个等待输入参数的函数，而 (Σg)(k) 是输入 k 的结果。按照这个符号，(Σg)(k) = k + C。将此函数输入 L − 1 算子可以重现 g。

To test the conjecture, apply the operator (L−1)Σ first to the easy function g = 1. Then Σg is a function waiting to be fed an argument, and (Σg)(k) is the result of feeding it k. With that notation, (Σg)(k) = k + C. Feeding this function to the L − 1 operator reproduces g.

使用次简单的函数（定义为 g(k) = k），不定和 (Σg)(k) 为 k(k − 1)/2 + C。将 Σg 再次经过 L − 1 可重现 g。

With the next-easiest function—defined by g(k) = k—the indefinite sum (Σg)(k) is k(k − 1)/2 + C. Passing Σg through L − 1 again reproduces g.

综上所述，对于测试函数 g(k) = 1 和 g(k) = k，算子乘积 (L − 1)Σ 将 g 取回自身，因此其作用类似于恒等算子。

In summary, for the test functions g(k) = 1 and g(k) = k, the operator product (L − 1)Σ takes g back to itself, so it acts like the identity operator.

这种行为具有普遍性——(L−1)Σ1 确实为 1，且 Σ= 1/(L−1)。由于 L = e ^D，因此有 Σ= 1/(e ^D − 1)。将泰勒级数的右侧展开，可以得到求和算符的惊人表示。

This behavior is general—(L−1)Σ1 is indeed 1, and Σ= 1/(L−1). Because L = e^D, we have Σ= 1/(e^D − 1). Expanding the right side in a Taylor series gives an amazing representation of the summation operator.

因为 D ∫ = 1，首项 1/D 是积分。因此，求和近似于积分——这是一个合理的结论，表明算子表示并非无稽之谈。

Because D ∫ = 1, the leading term 1/D is integration. Thus, summation is approximately integration—a plausible conclusion indicating that the operator representation is not nonsense.

将此算子系列应用于函数 f，然后在极限 a 和 b 处求值，得到欧拉-麦克劳林求和公式

Applying this operator series to a function f and then evaluating at the limits a and b produces the Euler–MacLaurin summation formula

其中 f ⁽ⁿ⁾表示 f 的 n 阶导数。

where f⁽ⁿ⁾ indicates the nth derivative of f.

和式中缺少通常的最后一个项 f(b)。包含这一项，可以得到一个有用的替代式

The sum lacks the usual final term f(b). Including this term gives the useful alternative

为了检验，尝试一个简单的例子：。使用欧拉-麦克劳林求和法，f(k) = k，a = 0，b = n。积分项贡献 n ² /2；常数项 [f(b) + f(a)]/2 贡献 n/2；其余项为零。结果熟悉且正确：

As a check, try an easy case: . Using Euler–MacLaurin summation, f(k) = k, a = 0, and b = n. The integral term then contributes n²/2; the constant term [f(b) + f(a)]/2 contributes n/2; and later terms vanish. The result is familiar and correct:

对欧拉-麦克劳林求和法更严格的检验是近似ln n!，即和k（见第4.5节）。因此，在（含）a = 1和b = n之间的和f(k) = ln k。结果为

A more stringent test of Euler–MacLaurin summation is to approximate ln n!, which is the sum k (Section 4.5). Therefore, sum f(k) = ln k between the (inclusive) limits a = 1 and b = n. The result is

1/D算子的积分贡献了ln k曲线下的面积。1/2算子的修正包含了三角形凸起（问题6.20）。省略号包含高阶修正（问题6.21）——用图片很难计算（问题4.32），但用欧拉-麦克劳林求和（问题6.21）很容易计算。

The integral, from the 1/D operator, contributes the area under the ln k curve. The correction, from the 1/2 operator, incorporates the triangular protrusions (Problem 6.20). The ellipsis includes the higher-order corrections (Problem 6.21)—hard to evaluate using pictures (Problem 4.32) but simple using Euler–MacLaurin summation (Problem 6.21).

6.4 正切根：一个令人畏惧的超越和

6.4 Tangent roots: A daunting transcendental sum

我们之所以选择这个告别的例子，是因为它的分析结合了多种街头斗争工具，它是一个难以解决的无限总和。

Our farewell example, chosen because its analysis combines diverse street-fighting tools, is a difficult infinite sum.

找出x _n是 tan x = x 的正解。

Find where the x_n are the positive solutions of tan x = x.

tan x = x 的解，或者等价地，tan x − x 的根，是超越函数，没有封闭形式，然而几乎每种求和法都需要封闭形式。街头斗争法会来帮我们。

The solutions to tan x = x or, equivalently, the roots of tan x − x, are transcendental and have no closed form, yet a closed form is required for almost every summation method. Street-fighting methods will come to our rescue.

6.4.1 图片和简单案例

6.4.1 Pictures and easy cases

从一个简单的案例开始分析。

Begin the analysis with a hopefully easy case.

x _{1 的}第一个根是多少？

What is the first root x₁?

tan x − x 的根由 y = x 与 y = tan x 的交点给出。令人惊讶的是，tan x 的分支中没有交点，当 0 < x < π/2 时（习题 6.23），第一个交点恰好位于 x = 3π/2 处的渐近线之前。因此，x ₁ ≈ 3π/2。

The roots of tan x − x are given by the intersections of y = x and y = tan x. Surprisingly, no intersection occurs in the branch of tan x where 0 < x < π/2 (Problem 6.23); the first intersection is just before the asymptote at x = 3π/2. Thus, x₁ ≈ 3π/2.

后续交叉路口大概在哪里？

Where, approximately, are the subsequent intersections?

_{随着 x 的增长，y = x 直线与 y = tan x 图像的交点越来越高，因此也越来越接近垂直渐近线。因此，对 x n}的较大部分进行以下渐近线近似：

As x grows, the y = x line intersects the y = tan x graph ever higher and therefore ever closer to the vertical asymptotes. Therefore, make the following asymptote approximation for the big part of x_n:

6.4.2 取出大部件

6.4.2 Taking out the big part

_{x n}的这个近似的低熵表达式给出了 S 的大部分（零近似值）。

This approximate, low-entropy expression for x_n gives the big part of S (the zeroth approximation).

从图片（第 4.5 节）或欧拉-麦克劳林求和（第 6.3.2 节）可以看出，该总和大致为以下积分。

The sum is, from a picture (Section 4.5) or from Euler–MacLaurin summation (Section 6.3.2), roughly the following integral.

所以，

Therefore,

阴影部分的凸起大致是三角形，它们的总面积是第一个矩形的一半。该矩形的面积为 1/9，所以

The shaded protrusions are roughly triangles, and they sum to one-half of the first rectangle. That rectangle has area 1/9, so

因此，S 的更准确估计是

Therefore, a more accurate estimate of S is

这比最初的估计略高。

which is slightly higher than the first estimate.

新的近似值是高估了还是低估了？

Is the new approximation an overestimate or an underestimate?

新的近似基于两个低估。首先，渐近线近似 x _n ≈ (n + 0.5)π 高估了每个 x _n，因此低估了和式中的倒数平方；其次，在进行渐近线近似后，和式的图形近似用内接三角形代替了每个凸起，从而低估了每个凸起（问题 6.24）。

The new approximation is based on two underestimates. First, the asymptote approximation x_n ≈ (n + 0.5)π overestimates each x_n and therefore underestimates the squared reciprocals in the sum Second, after making the asymptote approximation, the pictorial approximation to the sum replaces each protrusion with an inscribed triangle and thereby underestimates each protrusion (Problem 6.24).

如何弥补这两个低估？

How can these two underestimates be remedied?

第二个低估（凸起部分）可以通过精确求和来消除。这个和式之所以不为人熟知，部分原因在于它的第一项是分数 1/9——它的任意性增加了和式的熵。加上 n = 0 项（即 1）和偶数平方倒数 1/(2n) ^²，就得到了一个简洁而熟悉的低熵和式。

The second underestimate (the protrusions) is eliminated by summing exactly. The sum is unfamiliar partly because its first term is the fraction 1/9—whose arbitrariness increases the entropy of the sum. Including the n = 0 term, which is 1, and the even squared reciprocals 1/(2n)² produces a compact and familiar lower-entropy sum.

最终的低熵和就是著名的巴塞尔和（高熵结果通常并不出名）。它的值为 B = π ² /6（问题 6.22）。

The final, low-entropy sum is the famous Basel sum (high-entropy results are not often famous). Its value is B = π²/6 (Problem 6.22).

知道B = π ² /6如何帮助计算原始和 ？

How does knowing B = π²/6 help evaluate the original sum ?

对原始和的主要修改是加入了偶数平方倒数。它们的和是 B/4。

The major modification from the original sum was to include the even squared reciprocals. Their sum is B/4.

第二个修改是加入n = 0项。因此，为了得到，需要调整Basel值B，方法是先减去B/4，然后再减去n = 0项。代入B = π ² /6后，结果为

The second modification was to include the n = 0 term. Thus, to obtain , adjust the Basel value B by subtracting B/4 and then the n = 0 term. The result, after substituting B = π²/6, is

基于 x _n的渐近线近似，此精确总和可得出以下 S 估计值。

This exact sum, based on the asymptote approximation for x_n, produces the following estimate of S.

通过扩展产品进行简化，得到

Simplifying by expanding the product gives

这是使用渐近线近似 x _n ≈ (n + 0.5)π 的第三个 S 估计值。综合起来，这些估计值如下

This estimate of S is the third that uses the asymptote approximation x_n ≈ (n + 0.5)π. Assembled together, the estimates are

因为第三个估计值包含了的精确值，所以 S 估计值中的任何剩余误差都必须属于渐近线近似值本身。

Because the third estimate incorporated the exact value of , any remaining error in the estimate of S must belong to the asymptote approximation itself.

对于哪一项， 渐近线近似最不准确？

For which term of is the asymptote approximation most inaccurate?

随着 x 的增长，x 和 tan x 的图形交点越来越接近垂直渐近线。因此，渐近线近似在 n = 1 时产生最大的绝对误差。由于 x ₁是最小根，因此 x _{n的分数误差相对于 x n}的绝对误差，在 n = 1 时更加集中。 的分数误差是 x _n分数误差的 −2 倍（见 5.3 节），在 n = 1 时同样集中。因为（分数误差乘以自身）在 n = 1 时是最大的。

As x grows, the graphs of x and tan x intersect ever closer to the vertical asymptote. Thus, the asymptote approximation makes its largest absolute error when n = 1. Because x₁ is the smallest root, the fractional error in x_n is, relative to the absolute error in x_n, even more concentrated at n = 1. The fractional error in , being −2 times the fractional error in x_n (Section 5.3), is equally concentrated at n = 1. Because is the (the fractional error times itself) is, by far, the largest at n = 1.

由于误差集中在 n = 1 处，S 估计值的最大改进来自于用更精确的值替换近似值 x _{1 = (n + 0.5)π。一种简单的数值方法是使用牛顿-拉夫森法进行逐次逼近（}问题 4.38）。要用这种方法求根，首先对 x 进行一个初始猜测，然后使用替换值反复改进它

With the error so concentrated at n = 1, the greatest improvement in the estimate of S comes from replacing the approximation x₁ = (n + 0.5)π with a more accurate value. A simple numerical approach is successive approximation using the Newton–Raphson method (Problem 4.38). To find a root with this method, make a starting guess x and repeatedly improve it using the replacement

当 x 的起始猜测值略低于 1.5π 处的第一个渐近线时，该过程迅速收敛到 x ₁ = 4.4934...

When the starting guess for x is slightly below the first asymptote at 1.5π, the procedure rapidly converges to x₁ = 4.4934 . . .

因此，为了改进基于渐近线近似的估计值 S ≈ 0.094715，减去其近似的第一项（其较大部分）并添加校正后的第一项。

Therefore, to improve the estimate S ≈ 0.094715, which was based on the asymptote approximation, subtract its approximate first term (its big part) and add the corrected first term.

此外，使用牛顿-拉夫森法进行改进，1/x ^{2 2}项得出 S ≈ 0.09978（问题 6.27）。因此，一个合理的猜测是

Using the Newton–Raphson method to refine, in addition, the 1/x²² term gives S ≈ 0.09978 (Problem 6.27). Therefore, a highly educated guess is

未知超越数的无穷和似乎既不是超越数，也不是无理数！这个简单而令人惊讶的有理数值得一个简单的解释。

The infinite sum of unknown transcendental numbers seems to be neither transcendental nor irrational! This simple and surprising rational number deserves a simple explanation.

6.4.3 与多项式的类比

6.4.3 Analogy with polynomials

如果方程 tan x − x = 0 只有几个闭式解就好了！这样，和 S 就很容易计算了。用一个有单根的多项式方程代替 tan x − x 就可以实现这个愿望。最简单的有趣多项式是二次多项式，所以可以用一个简单的二次方程来做实验，例如 x ² − 3x + 2。

If only the equation tan x − x = 0 had just a few closed-form solutions! Then the sum S would be easy to compute. That wish is fulfilled by replacing tan x − x with a polynomial equation with simple roots. The simplest interesting polynomial is the quadratic, so experiment with a simple quadratic—for example, x² − 3x + 2.

该多项式有两个根，x ₁ = 1 和 x ₂ = 2；因此，与正切根和类似的多项式根和有两个项。

This polynomial has two roots, x₁ = 1 and x₂ = 2; therefore , the polynomial-root sum analog of the tangent-root sum, has two terms.

这种计算根和的蛮力方法需要二次方程的解。然而，如果方法可以转化为没有闭式解的方程 tan x − x = 0，就不能使用根本身。它只能利用二次方程的表面特征——即它的两个系数 2 和 −3。遗憾的是，没有一种可行的将 2 和 −3 结合起来的方法能够预测。

This brute-force method for computing the root sum requires a solution to the quadratic equation. However, a method that can transfer to the equation tan x − x = 0, which has no closed-form solution, cannot use the roots themselves. It must use only surface features of the quadratic—namely, its two coefficients 2 and −3. Unfortunately, no plausible method of combining 2 and −3 predicts that .

多项式类比哪里出错了？

Where did the polynomial analogy go wrong?

问题在于，二次函数 x ² − 3x + 2 与 tan x − x 不够相似。二次函数只有正根；然而，奇函数 tan x − x 有对称的正根和负根，并且在 x = 0 处有一个根。事实上，tan x 的泰勒级数为 x + x ³ /3 + 2x ⁵ /15 + · · ·（问题 6.28）；因此，

The problem is that the quadratic x² − 3x + 2 is not sufficiently similar to tan x − x. The quadratic has only positive roots; however, tan x − x, an odd function, has symmetric positive and negative roots and has a root at x = 0. Indeed, the Taylor series for tan x is x + x³/3 + 2x⁵/15 + · · · (Problem 6.28); therefore,

^{x 3}的公因数意味着 tan x − x 在 x = 0 处有一个三重根。类似的多项式——这里在 x = 0 处有一个三重根，一个正根和一个对称负根——是 (x+2)x ³ (x−2)，或者展开后为 x ⁵ − 4x ³。和式（使用正根）只包含一项。并且正好是 1/4。这个值可能是多项式最后两个系数的（负）比值。

The common factor of x³ means that tan x − x has a triple root at x = 0. An analogous polynomial—here, one with a triple root at x = 0, a positive root, and a symmetric negative root—is (x+2)x³(x−2) or, after expansion, x⁵ − 4x³. The sum (using the positive root) contains only one term and is simply 1/4. This value could plausibly arise as the (negative) ratio of the last two coefficients of the polynomial.

要判断该模式是否是巧合，可以尝试一个更复杂的多项式：其根为 −2、−1、0（三重）、1 和 2。其中一个这样的多项式是

To decide whether that pattern is a coincidence, try a richer polynomial: one with roots at −2, −1, 0 (threefold), 1, and 2. One such polynomial is

多项式根和仅使用两个正根 1 和 2，即 1/1 ² + 1/2 ²，等于 5/4——最后两个系数的（负）比。作为对此模式的最后测试，在根中包含 −3 和 3。得到的多项式为

The polynomial-root sum uses only the two positive roots 1 and 2 and is 1/1² + 1/2², which is 5/4—the (negative) ratio of the last two coefficients. As a final test of this pattern, include −3 and 3 among the roots. The resulting polynomial is

多项式根和使用三个正根 1、2 和 3，即 1/1 ² + 1/2 ² + 1/3 ²，即 49/36 — 同样是展开多项式中最后两个系数的（负）比率。

The polynomial-root sum uses the three positive roots 1, 2, and 3 and is 1/1² + 1/2² + 1/3², which is 49/36—again the (negative) ratio of the last two coefficients in the expanded polynomial.

这个模式的起源是什么？如何将其扩展到tan x − x ？

What is the origin of the pattern, and how can it be extended to tan x − x?

为了解释这个模式，将多项式整理如下：

To explain the pattern, tidy the polynomial as follows:

在这个排列中，和 49/36 是第一个值得关注的系数的负数。让我们推广一下。在 x = 0 处取 k 个根，在± x ₁ , ± x ₂ , . . ., ± x _n处取单个根，得到多项式

In this arrangement, the sum 49/36 appears as the negative of the first interesting coefficient. Let’s generalize. Placing k roots at x = 0 and single roots at ±x₁, ±x₂, . . ., ±x_n gives the polynomial

其中 A 为常数。当展开括号内因子的乘积时，展开式中x ²项的系数会从因子中的每个 x ² /x ² _{k项中获得一个贡献。因此，展开式开始于}

where A is a constant. When expanding the product of the factors in parentheses, the coefficient of the x² term in the expansion receives one contribution from each x²/x²_k term in a factor. Thus, the expansion begins

括号中的 x ²的系数是正切根和的多项式类比。

The coefficient of x² in parentheses is which is the polynomial analog of the tangent-root sum.

让我们将这种方法应用于 tan x − x。虽然它不是多项式，但它的泰勒级数就像一个无限次多项式。泰勒级数是

Let’s apply this method to tan x − x. Although it is not a polynomial, its Taylor series is like an infinite-degree polynomial. The Taylor series is

^{x 2}系数的负数应为 − 。因此，对于正切和问题，应为 −2/5 。不幸的是，正量的和不能为负！

The negative of the x² coefficient should be −. For the tangent-sum problem, should therefore be −2/5. Unfortunately, the sum of positive quantities cannot be negative!

这个类比哪里出了问题？

What went wrong with the analogy?

一个问题是，tan x − x 可能有虚数根或复数根，这些根的平方对 S 的贡献为负。幸运的是，它的所有根都是实数（问题 6.29）。一个更难解决的问题是，tan x − x 在有限的 x 值处趋于无穷大，并且无限次地这样做，而没有一个多项式会这样做，哪怕只有一次。

One problem is that tan x − x might have imaginary or complex roots whose squares contribute negative amounts to S. Fortunately, all its roots are real (Problem 6.29). A harder-to-solve problem is that tan x − x goes to infinity at finite values of x, and does so infinitely often, whereas no polynomial does so even once.

解决方法是构造一个没有无穷大但根与 tan x-x 相同的函数。tan x - x 的无穷大出现在 tan x 爆炸的地方，也就是 cos x = 0 的地方。为了在不产生或消除任何根的情况下消除无穷大，将 tan x - x 乘以 cos x。因此，要展开的类多项式函数是 sin x - x cos x。

The solution is to construct a function having no infinities but having the same roots as tan x−x. The infinities of tan x − x occur where tan x blows up, which is where cos x = 0. To remove the infinities without creating or destroying any roots, multiply tan x − x by cos x. The polynomial-like function to expand is therefore sin x − x cos x.

其泰勒展开式为

Its Taylor expansion is

这两个系列的区别在于

The difference of the two series is

x ³ /3 因子表示 x = 0 处的三重根。最后，作为 x ²系数的负数，就是我们的正切根和 S = 1/10。

The x³/3 factor indicates the triple root at x = 0. And there at last, as the negative of the x² coefficient, sits our tangent-root sum S = 1/10.

6.5 一路顺风

6.5 Bon voyage

希望您喜欢将街头格斗技巧融入您的问题解决工具箱。希望您能找到各种机会运用量纲分析、简单案例、归并法、图像推理、去重法和类比法。随着您运用这些工具，您会不断磨练它们，甚至创造出新的工具。

I hope that you have enjoyed incorporating street-fighting methods into your problem-solving toolbox. May you find diverse opportunities to use dimensional analysis, easy cases, lumping, pictorial reasoning, taking out the big part, and analogy. As you apply the tools, you will sharpen them—and even build new tools.